Q18: For what value of λ is the function defined by f(x)= continuous at x=0? What about continuity at x=1?

Given: The function f is defined by f(x)= To Find: 1. The value of λ for which the function is continuous at x=0. 2. The continuity of the function at x=1. Solution: Part 1: Continuity at x=0 For the function to be continuous at x=0, we must have lim_x → 0^- f(x) = lim_x → 0^+ f(x) = f(0).…

Continuity and Differentiability Class 12 Mathematics NCERT Solutions

Q1EXERCISE 5.1

Prove that the function $f(x)=5 x-3$ is continuous at $x=0$ , at $x=-3$ and at $x=5$ .

Solution

Given: The function

f(x) = 5x - 3

.

To Prove: The function

f(x)

is continuous at

x=0

, at

x=-3

, and at

x=5

.

A function is continuous at a point

x=c

if

\lim_{x \to c} f(x) = f(c)

.

Case 1: Continuity at $x=0$

Value of the function at $x=0$ : $f(0) = 5(0) - 3 = 0 - 3 = -3$ .
Limit of the function as $x$ approaches 0: $\lim_{x \to 0} f(x) = \lim_{x \to 0} (5x - 3) = 5(0) - 3 = -3$
Comparison: Since $\lim_{x \to 0} f(x) = f(0) = -3$ , the function is continuous at $x=0$ .

Case 2: Continuity at $x=-3$

Value of the function at $x=-3$ : $f(-3) = 5(-3) - 3 = -15 - 3 = -18$ .
Limit of the function as $x$ approaches -3: $\lim_{x \to -3} f(x) = \lim_{x \to -3} (5x - 3) = 5(-3) - 3 = -15 - 3 = -18$
Comparison: Since $\lim_{x \to -3} f(x) = f(-3) = -18$ , the function is continuous at $x=-3$ .

Case 3: Continuity at $x=5$

Value of the function at $x=5$ : $f(5) = 5(5) - 3 = 25 - 3 = 22$ .
Limit of the function as $x$ approaches 5: $\lim_{x \to 5} f(x) = \lim_{x \to 5} (5x - 3) = 5(5) - 3 = 25 - 3 = 22$
Comparison: Since $\lim_{x \to 5} f(x) = f(5) = 22$ , the function is continuous at $x=5$ .

Final Answer: Thus, it is proved that the function

f(x)=5x-3

is continuous at

x=0

, at

x=-3

, and at

x=5

.

Q2EXERCISE 5.1

Examine the continuity of the function $f(x)=2 x^{2}-1$ at $x=3$ .

Solution

Given: The function

f(x) = 2x^2 - 1

.

To Find: Examine the continuity of the function at

x=3

.

A function is continuous at a point

x=c

if

\lim_{x \to c} f(x) = f(c)

. Here,

c=3

.

Solution:

Value of the function at $x=3$ : $f(3) = 2(3)^2 - 1 = 2(9) - 1 = 18 - 1 = 17$ .
Limit of the function as $x$ approaches 3: Since $f(x)$ is a polynomial function, the limit can be found by direct substitution. $\lim_{x \to 3} f(x) = \lim_{x \to 3} (2x^2 - 1) = 2(3)^2 - 1 = 17$
Comparison: We observe that $\lim_{x \to 3} f(x) = 17$ and $f(3) = 17$ . Since $\lim_{x \to 3} f(x) = f(3)$ , the function is continuous at $x=3$ .

Final Answer: The function

f(x) = 2x^2 - 1

is continuous at

x=3

.

Q3EXERCISE 5.1

Examine the following functions for continuity.

(a)

f(x)=x-5

(b)

f(x)=\frac{1}{x-5}, x \neq 5

(c)

f(x)=\frac{x^{2}-25}{x+5}, x \neq -5

(d)

f(x)=|x-5|

Solution

To Find: Examine the following functions for continuity.

(a) $f(x) = x - 5$

Solution: The function is

f(x) = x - 5

. This is a polynomial function. The domain of

f

is the set of all real numbers,

\mathbb{R}

. Let

c

be any arbitrary real number.

Value of the function at $x=c$ : $f(c) = c - 5$ .
Limit of the function as $x$ approaches $c$ : $\lim_{x \to c} f(x) = \lim_{x \to c} (x - 5) = c - 5$

Since

\lim_{x \to c} f(x) = f(c)

for any real number

c

, the function

f(x) = x-5

is continuous for all

x \in \mathbb{R}

.

(b) $f(x) = \frac{1}{x-5}, x \neq 5$

Solution: The function is

f(x) = \frac{1}{x-5}

. The domain of

f

is all real numbers except 5, i.e.,

x \in \mathbb{R} - \{5\}

. Let

c

be any real number in the domain, so

c \neq 5

.

Value of the function at $x=c$ : $f(c) = \frac{1}{c-5}$ .
Limit of the function as $x$ approaches $c$ : $\lim_{x \to c} f(x) = \lim_{x \to c} \frac{1}{x-5} = \frac{1}{c-5}$

Since

\lim_{x \to c} f(x) = f(c)

for any

c

in its domain, the function

f(x) = \frac{1}{x-5}

is continuous on its domain.

(c) $f(x) = \frac{x^2 - 25}{x+5}, x \neq -5$

Solution: The function is

f(x) = \frac{x^2 - 25}{x+5}

. The domain of

f

is all real numbers except -5, i.e.,

x \in \mathbb{R} - \{-5\}

. For any

x

in the domain, we can simplify the function:

f(x) = \frac{(x-5)(x+5)}{x+5} = x-5

Let

c

be any real number in the domain, so

c \neq -5

.

Value of the function at $x=c$ : $f(c) = c-5$ .
Limit of the function as $x$ approaches $c$ : $\lim_{x \to c} f(x) = \lim_{x \to c} (x-5) = c-5$

Since

\lim_{x \to c} f(x) = f(c)

for any

c

in its domain, the function

f(x) = \frac{x^2 - 25}{x+5}

is continuous on its domain.

(d) $f(x) = |x - 5|$

Solution: The function is

f(x) = |x - 5|

. This can be written as a piecewise function:

f(x) = \begin{cases} x-5, & \text{if } x-5 \geq 0 \implies x \geq 5 \ -(x-5), & \text{if } x-5 < 0 \implies x < 5 \end{cases}

The domain of

f

is all real numbers. We check for continuity at the point

x=5

where the definition changes. For all other points, the function is a polynomial and hence continuous.

Continuity at $x=5$ :

Left-Hand Limit (LHL): $\lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} -(x-5) = -(5-5) = 0$
Right-Hand Limit (RHL): $\lim_{x \to 5^+} f(x) = \lim_{x \to 5^+} (x-5) = 5-5 = 0$
Value of the function at $x=5$ : $f(5) = |5-5| = 0$ .

Since LHL = RHL =

f(5)

, the function is continuous at

x=5

. As the function is also continuous for all

x \neq 5

, it is continuous for all real numbers.

Final Answer: (a) Continuous for all

x \in \mathbb{R}

. (b) Continuous on its domain,

\mathbb{R} - \{5\}

. (c) Continuous on its domain,

\mathbb{R} - \{-5\}

. (d) Continuous for all

x \in \mathbb{R}

.

Q4EXERCISE 5.1

Prove that the function $f(x)=x^{n}$ is continuous at $x=n$ , where $n$ is a positive integer.

Solution

Given: The function

f(x) = x^n

, where

n

is a positive integer.

To Prove: The function

f(x)

is continuous at

x=n

.

A function is continuous at a point

x=c

if

\lim_{x \to c} f(x) = f(c)

. Here, the point is

c=n

.

Proof:

Value of the function at $x=n$ : $f(n) = n^n$ .
Limit of the function as $x$ approaches $n$ : The function $f(x) = x^n$ is a polynomial function for a positive integer $n$ . The limit of a polynomial function at a point can be found by direct substitution. $\lim_{x \to n} f(x) = \lim_{x \to n} (x^n)$ $= n^n$
Comparison: We have $\lim_{x \to n} f(x) = n^n$ and $f(n) = n^n$ . Since $\lim_{x \to n} f(x) = f(n)$ , the function is continuous at $x=n$ .

Hence Proved.

Final Answer: The function

f(x)=x^n

is continuous at

x=n

, where

n

is a positive integer.

Q5EXERCISE 5.1

Is the function $f$ defined by $x, \text { if } x \leq 1 \ 5, \text { if } x>1 \end{array}\right.$$ continuous at $x=0$ ? At $x=1$ ? At $x=2$ ?$

Solution

Given: The function

f(x)

defined as:

f(x)=\left\{\begin{array}{ll} x, & \text { if } x \leq 1 \ 5, & \text { if } x > 1 \end{array}\right.

To Find: Whether the function is continuous at

x=0

, at

x=1

, and at

x=2

.

Case 1: Continuity at $x=0$

For

x=0

, the condition

x \leq 1

is met. So, for

x

in the neighborhood of 0, we use

f(x)=x

.

Value of the function at $x=0$ : $f(0) = 0$ .
Limit of the function as $x$ approaches 0: $\lim_{x \to 0} f(x) = \lim_{x \to 0} x = 0$
Comparison: Since $\lim_{x \to 0} f(x) = f(0)$ , the function is continuous at $x=0$ .

Case 2: Continuity at $x=1$

The point

x=1

is where the function definition changes. We must check the left-hand limit (LHL) and the right-hand limit (RHL).

Left-Hand Limit (LHL): For $x \to 1^-$ , we have $x < 1$ , so we use $f(x)=x$ . $LHL = \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x = 1$
Right-Hand Limit (RHL): For $x \to 1^+$ , we have $x > 1$ , so we use $f(x)=5$ . $RHL = \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} 5 = 5$
Comparison: Since LHL $\neq$ RHL (because $1 \neq 5$ ), the limit $\lim_{x \to 1} f(x)$ does not exist. Therefore, the function is not continuous at $x=1$ .

Case 3: Continuity at $x=2$

For

x=2

, the condition

x > 1

is met. So, for

x

in the neighborhood of 2, we use

f(x)=5

.

Value of the function at $x=2$ : $f(2) = 5$ .
Limit of the function as $x$ approaches 2: $\lim_{x \to 2} f(x) = \lim_{x \to 2} 5 = 5$
Comparison: Since $\lim_{x \to 2} f(x) = f(2)$ , the function is continuous at $x=2$ .

Final Answer:

The function $f$ is continuous at $x=0$ .
The function $f$ is not continuous at $x=1$ .
The function $f$ is continuous at $x=2$ .

Q6EXERCISE 5.1

Find all points of discontinuity of $f$ , where $f$ is defined by 6. $f(x)=\left\{\begin{array}{l} 2 x+3, \text { if } x \leq 2 \ 2 x-3, \text { if } x>2 \end{array}\right.$

Solution

Given: The function

f

is defined by

f(x)=\left\{\begin{array}{ll} 2 x+3, & \text { if } x \leq 2 \ 2 x-3, & \text { if } x>2 \end{array}\right.

To Find: All points of discontinuity of

f

.

Solution: The function

f

is defined on all points of the real line. Let's consider a point

c

on the real line.

Case 1: $c < 2$ In this case,

f(x) = 2x+3

. Since this is a polynomial function, it is continuous for all

x < 2

. So,

f

is continuous for all

x=c < 2

.

Case 2: $c > 2$ In this case,

f(x) = 2x-3

. Since this is a polynomial function, it is continuous for all

x > 2

. So,

f

is continuous for all

x=c > 2

.

Case 3: $c = 2$ We check the continuity at the point

x=2

. For

f

to be continuous at

x=2

, we must have

\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2)

.

Left-Hand Limit (LHL):

\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (2x+3) = 2(2) + 3 = 4+3 = 7

Right-Hand Limit (RHL):

\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (2x-3) = 2(2) - 3 = 4-3 = 1

Value of the function at $x=2$ :

f(2) = 2(2) + 3 = 4+3 = 7

Since the Left-Hand Limit (

7

) is not equal to the Right-Hand Limit (

1

), the limit

\lim_{x \to 2} f(x)

does not exist. Therefore, the function

f

is discontinuous at

x=2

.

Final Answer: The only point of discontinuity for the function

f

is

x=2

.

Q7EXERCISE 5.1

Find all points of discontinuity of $f$ , where $f$ is defined by 7. $f(x)= \begin{cases}|x|+3, & \text { if } x \leq -3 \ -2 x, & \text { if }-3<x<3 \ 6 x+2, & \text { if } x \geq 3\end{cases}$

Solution

Given: The function

f

is defined by $$f(x)= \begin{cases}|x|+3, & \text { if } x \leq -3 \ -2 x, & \text { if }-3 3

,

f(x)=6x+2$ is continuous.

Final Answer: The only point of discontinuity for the function

f

is

x=3

.

Q8EXERCISE 5.1

Find all points of discontinuity of $f$ , where $f$ is defined by 8. $f(x)=\left\{\begin{array}{cc} \frac{|x|}{x}, & \text { if } x \neq 0 \ 0, & \text { if } x=0 \end{array}\right.$

Solution

Given: The function

f

is defined by

f(x)=\left\{\begin{array}{ll} \frac{|x|}{x}, & \text { if } x \neq 0 \ 0, & \text { if } x=0 \end{array}\right.

To Find: All points of discontinuity of

f

.

Solution: The function

f

is defined for all real numbers. The only potential point of discontinuity is at

x=0

. Let's analyze the function definition for

x \neq 0

.

If

x > 0

, then

|x|=x

. So,

f(x) = \frac{x}{x} = 1

. If

x < 0

, then

|x|=-x

. So,

f(x) = \frac{-x}{x} = -1

.

So the function can be rewritten as:

f(x)=\left\{\begin{array}{cl} 1, & \text { if } x>0 \ -1, & \text { if } x<0 \ 0, & \text { if } x=0 \end{array}\right.

Now, we check the continuity at

x=0

. Left-Hand Limit (LHL):

\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-1) = -1

Right-Hand Limit (RHL):

\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (1) = 1

Value of the function at $x=0$ :

f(0) = 0

Since LHL

\neq

RHL, the limit

\lim_{x \to 0} f(x)

does not exist. Therefore, the function

f

is discontinuous at

x=0

.

For any

c \neq 0

, the function is a constant function (

f(x)=1

for

c>0

or

f(x)=-1

for

c<0

), which is continuous everywhere in its domain. Thus,

f

is continuous for all

x \neq 0

.

Final Answer: The only point of discontinuity for the function

f

is

x=0

.

Q9EXERCISE 5.1

Find all points of discontinuity of $f$ , where $f$ is defined by 9. $f(x)= \begin{cases}\frac{x}{|x|}, & \text { if } x<0 \ -1, & \text { if } x \geq 0\end{cases}$

Solution

Given: The function

f

is defined by

f(x)= \begin{cases}\frac{x}{|x|}, & \text { if } x<0 \ -1, & \text { if } x \geq 0\end{cases}

To Find: All points of discontinuity of

f

.

Solution: The function

f

is defined for all real numbers. Let's analyze the definition for

x<0

. If

x<0

, then

|x| = -x

. So, for

x<0

,

f(x) = \frac{x}{-x} = -1

.

The function definition can be simplified to:

f(x)= \begin{cases}-1, & \text { if } x<0 \ -1, & \text { if } x \geq 0\end{cases}

This means that

f(x) = -1

for all real numbers

x \in \mathbb{R}

.

Since

f(x)

is a constant function, it is continuous for all real numbers.

To be formal, let's check continuity at the point

x=0

where the definition changes. Left-Hand Limit (LHL):

\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-1) = -1

Right-Hand Limit (RHL):

\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (-1) = -1

Value of the function at $x=0$ :

f(0) = -1

Since LHL = RHL =

f(0)

, the function is continuous at

x=0

. Since the function is a constant function, it is continuous everywhere.

Final Answer: There is no point of discontinuity for the function

f

.

Q10EXERCISE 5.1

Find all points of discontinuity of $f$ , where $f$ is defined by 10. $f(x)=\left\{\begin{array}{l} x+1, \text { if } x \geq 1 \ x^{2}+1, \text { if } x<1 \end{array}\right.$

Solution

Given: The function

f

is defined by

f(x)=\left\{\begin{array}{ll} x+1, & \text { if } x \geq 1 \ x^{2}+1, & \text { if } x<1 \end{array}\right.

To Find: All points of discontinuity of

f

.

Solution: The function

f

is defined on all points of the real line. The only point where continuity needs to be checked is

x=1

.

Case 1: $c < 1$ In this case,

f(x) = x^2+1

. Since this is a polynomial function, it is continuous for all

x < 1

. So,

f

is continuous for all

x=c < 1

.

Case 2: $c > 1$ In this case,

f(x) = x+1

. Since this is a polynomial function, it is continuous for all

x > 1

. So,

f

is continuous for all

x=c > 1

.

Case 3: $c = 1$ We check the continuity at the point

x=1

. For

f

to be continuous at

x=1

, we must have

\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)

.

Left-Hand Limit (LHL):

\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^2+1) = (1)^2 + 1 = 1+1 = 2

Right-Hand Limit (RHL):

\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x+1) = 1 + 1 = 2

Value of the function at $x=1$ :

f(1) = 1 + 1 = 2

Since the Left-Hand Limit = Right-Hand Limit = Value of the function at

x=1

, i.e.,

\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) = 2

Therefore, the function

f

is continuous at

x=1

.

As the function is continuous for

x<1

,

x>1

, and also at

x=1

, it is continuous for all real numbers.

Final Answer: There is no point of discontinuity for the function

f

.

Q11EXERCISE 5.1

Find all points of discontinuity of $f$ , where $f$ is defined by 11. $f(x)= \begin{cases}x^{3}-3, & \text { if } x \leq 2 \ x^{2}+1, & \text { if } x>2\end{cases}$

Solution

Given: The function is defined as

f(x)= \begin{cases}x^{3}-3, & \text { if } x \leq 2 \ x^{2}+1, & \text { if } x>2\end{cases}

.

To Find: All points of discontinuity of the function

f

.

Solution: The function

f

is defined on all real numbers. The continuity of

f

can be checked in three cases.

Case 1: For any real number $c < 2$ In this case,

f(x) = x^3 - 3

. Since this is a polynomial function, it is continuous for all

x < 2

. Thus,

f

is continuous for all real numbers

c < 2

.

Case 2: For any real number $c > 2$ In this case,

f(x) = x^2 + 1

. Since this is a polynomial function, it is continuous for all

x > 2

. Thus,

f

is continuous for all real numbers

c > 2

.

Case 3: At $x = 2$ We need to check the continuity at the point

x=2

where the function definition changes.

Value of the function at

x=2

:

f(2) = 2^3 - 3 = 8 - 3 = 5

.

Left-Hand Limit (LHL) at

x=2

:

\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x^3 - 3) = 2^3 - 3 = 8 - 3 = 5

Right-Hand Limit (RHL) at

x=2

:

\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (x^2 + 1) = 2^2 + 1 = 4 + 1 = 5

Since the Left-Hand Limit, Right-Hand Limit, and the value of the function at

x=2

are all equal:

\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2) = 5

So, the function

f

is continuous at

x=2

.

From all three cases, we can conclude that the function

f

is continuous for all real numbers.

Final Answer: There are no points of discontinuity for the given function.

Q12EXERCISE 5.1

Find all points of discontinuity of $f$ , where $f$ is defined by 12. $f(x)= \begin{cases}x^{10}-1, & \text { if } x \leq 1 \ x^{2}, & \text { if } x>1\end{cases}$

Solution

Given: The function is defined as

f(x)= \begin{cases}x^{10}-1, & \text { if } x \leq 1 \ x^{2}, & \text { if } x>1\end{cases}

.

To Find: All points of discontinuity of the function

f

.

Solution: The function

f

is defined on all real numbers. The continuity of

f

can be checked in three cases.

Case 1: For any real number $c < 1$ In this case,

f(x) = x^{10} - 1

. Since this is a polynomial function, it is continuous for all

x < 1

. Thus,

f

is continuous for all real numbers

c < 1

.

Case 2: For any real number $c > 1$ In this case,

f(x) = x^2

. Since this is a polynomial function, it is continuous for all

x > 1

. Thus,

f

is continuous for all real numbers

c > 1

.

Case 3: At $x = 1$ We need to check the continuity at the point

x=1

where the function definition changes.

Value of the function at

x=1

:

f(1) = 1^{10} - 1 = 1 - 1 = 0

.

Left-Hand Limit (LHL) at

x=1

:

\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^{10} - 1) = 1^{10} - 1 = 1 - 1 = 0

Right-Hand Limit (RHL) at

x=1

:

\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x^2) = 1^2 = 1

Since the Left-Hand Limit is not equal to the Right-Hand Limit:

\lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x)

So, the function

f

is not continuous at

x=1

.

From all three cases, we can conclude that the only point of discontinuity is

x=1

.

Final Answer: The only point of discontinuity for the given function is

x=1

.

Q13EXERCISE 5.1

Is the function defined by $f(x)= \begin{cases}x+5, & \text { if } x \leq 1 \ x-5, & \text { if } x>1\end{cases}$ a continuous function?

Solution

Given: The function is defined as

f(x)= \begin{cases}x+5, & \text { if } x \leq 1 \ x-5, & \text { if } x>1\end{cases}

.

To determine: Whether the given function is a continuous function.

Solution: A function is continuous if it is continuous at every point in its domain. The domain of the given function is the set of all real numbers,

\mathbb{R}

. The function's definition changes at

x=1

, so we must check for continuity at this point.

We can analyze the continuity in three cases:

Case 1: For any real number $c < 1$ In this case,

f(x) = x+5

. This is a linear (polynomial) function, which is continuous everywhere. So,

f

is continuous for all

x < 1

.

Case 2: For any real number $c > 1$ In this case,

f(x) = x-5

. This is also a linear (polynomial) function, which is continuous everywhere. So,

f

is continuous for all

x > 1

.

Case 3: At $x = 1$ We check the continuity at

x=1

.

Value of the function at

x=1

:

f(1) = 1+5 = 6

.

Left-Hand Limit (LHL) at

x=1

:

\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x+5) = 1+5 = 6

Right-Hand Limit (RHL) at

x=1

:

\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x-5) = 1-5 = -4

Here, the Left-Hand Limit is not equal to the Right-Hand Limit:

\lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x)

Therefore, the function

f

is not continuous at

x=1

.

Since the function is not continuous at

x=1

, it is not a continuous function over its entire domain.

Final Answer: No, the function is not a continuous function. It is discontinuous at

x=1

.

Q14EXERCISE 5.1

Discuss the continuity of the function $f$ , where $f$ is defined by $f(x)=\left\{\begin{array}{l}3, \text { if } 0 \leq x \leq 1 \ 4, \text { if } 1<x<3 \ 5, \text { if } 3 \leq x \leq 10\end{array}\right.$

Solution

Given: The function is defined as

f(x)=\left\{\begin{array}{l}3, \text { if } 0 \leq x \leq 1 \ 4, \text { if } 1<x<3 \ 5, \text { if } 3 \leq x \leq 10\end{array}\right.

.

To Find: Discuss the continuity of the function

f

.

Solution: The domain of the function is the interval

[0, 10]

. The function definition changes at

x=1

and

x=3

. We need to check the continuity at these points.

Case 1: In the open intervals $(0, 1)$ , $(1, 3)$ , and $(3, 10)$

For $x \in (0, 1)$ , $f(x) = 3$ , which is a constant function and hence continuous.
For $x \in (1, 3)$ , $f(x) = 4$ , which is a constant function and hence continuous.
For $x \in (3, 10)$ , $f(x) = 5$ , which is a constant function and hence continuous.

Case 2: At $x = 1$ Value of the function at

x=1

:

f(1) = 3

.

Left-Hand Limit (LHL) at

x=1

:

\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (3) = 3

Right-Hand Limit (RHL) at

x=1

:

\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (4) = 4

Since LHL

\neq

RHL (

3 \neq 4

), the function

f

is discontinuous at

x=1

.

Case 3: At $x = 3$ Value of the function at

x=3

:

f(3) = 5

.

Left-Hand Limit (LHL) at

x=3

:

\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (4) = 4

Right-Hand Limit (RHL) at

x=3

:

\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (5) = 5

Since LHL

\neq

RHL (

4 \neq 5

), the function

f

is discontinuous at

x=3

.

Also, at the endpoints of the domain: At

x=0

,

f(0)=3

.

\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (3) = 3

. So

f

is right-continuous at

x=0

. At

x=10

,

f(10)=5

.

\lim_{x \to 10^-} f(x) = \lim_{x \to 10^-} (5) = 5

. So

f

is left-continuous at

x=10

.

Final Answer: The function

f

is continuous on its domain

[0, 10]

except at the points

x=1

and

x=3

. The points of discontinuity are

x=1

and

x=3

.

Q15EXERCISE 5.1

Discuss the continuity of the function $f$ , where $f$ is defined by $f(x)= \begin{cases}2 x, & \text { if } x<0 \ 0, & \text { if } 0 \leq x \leq 1 \ 4 x, & \text { if } x>1\end{cases}$

Solution

Given: The function is defined as

f(x)= \begin{cases}2 x, & \text { if } x<0 \ 0, & \text { if } 0 \leq x \leq 1 \ 4 x, & \text { if } x>1\end{cases}

.

To Find: Discuss the continuity of the function

f

.

Solution: The domain of the function is the set of all real numbers,

\mathbb{R}

. The function definition changes at

x=0

and

x=1

. We need to check the continuity at these points.

Case 1: In the open intervals $(-\infty, 0)$ , $(0, 1)$ , and $(1, \infty)$

For $x < 0$ , $f(x) = 2x$ , which is a polynomial function and hence continuous.
For $0 < x < 1$ , $f(x) = 0$ , which is a constant function and hence continuous.
For $x > 1$ , $f(x) = 4x$ , which is a polynomial function and hence continuous.

Case 2: At $x = 0$ Value of the function at

x=0

:

f(0) = 0

.

Left-Hand Limit (LHL) at

x=0

:

\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (2x) = 2(0) = 0

Right-Hand Limit (RHL) at

x=0

:

\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (0) = 0

Since LHL = RHL =

f(0) = 0

, the function

f

is continuous at

x=0

.

Case 3: At $x = 1$ Value of the function at

x=1

:

f(1) = 0

.

Left-Hand Limit (LHL) at

x=1

:

\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (0) = 0

Right-Hand Limit (RHL) at

x=1

:

\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (4x) = 4(1) = 4

Since LHL

\neq

RHL (

0 \neq 4

), the function

f

is discontinuous at

x=1

.

Final Answer: The function

f

is continuous for all real numbers except at

x=1

. The only point of discontinuity is

x=1

.

Q16EXERCISE 5.1

Discuss the continuity of the function $f$ , where $f$ is defined by $f(x)= \begin{cases}-2, & \text { if } x \leq -1 \ 2 x, & \text { if }-11\end{cases}$

Solution

Given: The function

f

is defined by

f(x)= \begin{cases}-2, & \text { if } x \leq -1 \ 2 x, & \text { if }-11\end{cases}

To Find: Discuss the continuity of the function

f(x)

.

Solution: The function

f(x)

is defined on the entire real line. The definition of the function changes at

x = -1

and

x = 1

. We need to check the continuity at these points.

For any

c 1

,

f(x) = 2

, which is a constant function. Thus,

f

is continuous for all

x > 1

.

Now, we check the continuity at

x = -1

. Left-Hand Limit (LHL) at

x=-1

:

\lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} (-2) = -2

Right-Hand Limit (RHL) at

x=-1

:

\lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} (2x) = 2(-1) = -2

Value of the function at

x=-1

:

f(-1) = -2

Since LHL = RHL =

f(-1)

, the function is continuous at

x = -1

.

Next, we check the continuity at

x = 1

. Left-Hand Limit (LHL) at

x=1

:

\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (2x) = 2(1) = 2

Right-Hand Limit (RHL) at

x=1

:

\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2) = 2

Value of the function at

x=1

:

f(1) = 2(1) = 2

Since LHL = RHL =

f(1)

, the function is continuous at

x = 1

.

Since the function is continuous at all points where its definition changes, and is also continuous at all other points in its domain, the function

f(x)

is continuous for all real numbers.

Final Answer: The function

f(x)

is continuous on the entire real line

\mathbb{R}

. There are no points of discontinuity.

Q17EXERCISE 5.1

Find the relationship between $a$ and $b$ so that the function $f$ defined by $f(x)= \begin{cases}a x+1, & \text { if } x \leq 3 \ b x+3, & \text { if } x>3\end{cases}$ is continuous at $x=3$ .

Solution

Given: The function

f

is defined by

f(x)= \begin{cases}a x+1, & \text { if } x \leq 3 \ b x+3, & \text { if } x>3\end{cases}

It is also given that the function is continuous at

x=3

.

To Find: The relationship between

a

and

b

.

Solution: For a function to be continuous at a point

x=c

, we must have:

\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)

Here,

c=3

. Since the function is continuous at

x=3

, we have:

\lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) = f(3)

First, we find the Left-Hand Limit (LHL):

\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (ax+1) = a(3) + 1 = 3a+1

Next, we find the Right-Hand Limit (RHL):

\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (bx+3) = b(3) + 3 = 3b+3

The value of the function at

x=3

is:

f(3) = a(3) + 1 = 3a+1

For continuity, LHL must be equal to RHL.

3a+1 = 3b+3

3a - 3b = 3 - 1

3a - 3b = 2

3(a - b) = 2

a - b = \frac{2}{3}

This can also be written as

a = b + \frac{2}{3}

.

Final Answer: The relationship between

a

and

b

is

a - b = \frac{2}{3}

or

a = b + \frac{2}{3}

.

Q18EXERCISE 5.1

For what value of $\lambda$ is the function defined by $f(x)= \begin{cases}\lambda\left(x^{2}-2 x\right), & \text { if } x \leq 0 \ 4 x+1, & \text { if } x>0\end{cases}$ continuous at $x=0$ ? What about continuity at $x=1$ ?

Solution

Given: The function

f

is defined by

f(x)= \begin{cases}\lambda\left(x^{2}-2 x\right), & \text { if } x \leq 0 \ 4 x+1, & \text { if } x>0\end{cases}

To Find:

The value of $\lambda$ for which the function is continuous at $x=0$ .
The continuity of the function at $x=1$ .

Solution:

Part 1: Continuity at $x=0$ For the function to be continuous at

x=0

, we must have

\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)

.

Left-Hand Limit (LHL) at

x=0

:

\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \lambda(x^2 - 2x) = \lambda(0^2 - 2(0)) = 0

Right-Hand Limit (RHL) at

x=0

:

\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (4x+1) = 4(0) + 1 = 1

Value of the function at

x=0

:

f(0) = \lambda(0^2 - 2(0)) = 0

For continuity, we need LHL = RHL. However, we found that LHL = 0 and RHL = 1. Since

0 \neq 1

, the left-hand limit is not equal to the right-hand limit. This is true regardless of the value of

\lambda

. Therefore, the function cannot be continuous at

x=0

for any value of

\lambda

.

Part 2: Continuity at $x=1$ For

x=1

, we have

x > 0

, so the function is defined as

f(x) = 4x+1

. This is a polynomial function, and polynomial functions are continuous everywhere. Therefore,

f(x)

is continuous at

x=1

.

To verify this using limits:

\lim_{x \to 1} f(x) = \lim_{x \to 1} (4x+1) = 4(1) + 1 = 5

And the value of the function at

x=1

is:

f(1) = 4(1) + 1 = 5

Since

\lim_{x \to 1} f(x) = f(1)

, the function is continuous at

x=1

. The continuity at

x=1

does not depend on the value of

\lambda

.

Final Answer: There is no value of

\lambda

for which the function is continuous at

x=0

. The function is continuous at

x=1

for any real value of

\lambda

.

Q19EXERCISE 5.1

Show that the function defined by $g(x)=x-[x]$ is discontinuous at all integral points. Here $[x]$ denotes the greatest integer less than or equal to $x$ .

Solution

Given: The function

g(x) = x - [x]

, where

[x]

denotes the greatest integer less than or equal to

x

.

To Prove: The function

g(x)

is discontinuous at all integral points.

Proof: Let

c

be an arbitrary integer. We will check the continuity of

g(x)

at the point

x=c

. For a function to be continuous at

x=c

, we must have

\lim_{x \to c^-} g(x) = \lim_{x \to c^+} g(x) = g(c)

.

First, let's calculate the value of the function at

x=c

: Since

c

is an integer, the greatest integer less than or equal to

c

is

c

itself. So,

[c] = c

.

g(c) = c - [c] = c - c = 0

Now, let's find the Left-Hand Limit (LHL) at

x=c

:

\lim_{x \to c^-} g(x) = \lim_{x \to c^-} (x - [x])

As

x

approaches

c

from the left,

x

is slightly less than

c

. For example,

x = c - h

where

h

is a small positive number. For any such

x

, the greatest integer less than or equal to

x

is

c-1

. So,

[x] = c-1

.

\text{LHL} = \lim_{x \to c^-} (x - (c-1)) = c - (c-1) = c - c + 1 = 1

Next, let's find the Right-Hand Limit (RHL) at

x=c

:

\lim_{x \to c^+} g(x) = \lim_{x \to c^+} (x - [x])

As

x

approaches

c

from the right,

x

is slightly greater than

c

. For example,

x = c + h

where

h

is a small positive number. For any such

x

, the greatest integer less than or equal to

x

is

c

. So,

[x] = c

.

\text{RHL} = \lim_{x \to c^+} (x - c) = c - c = 0

Comparing the limits and the function value: LHL = 1 RHL = 0

g(c) = 0

Since LHL

\neq

RHL, the limit of

g(x)

as

x

approaches

c

does not exist. Therefore, the function

g(x)

is not continuous at

x=c

.

Since

c

was chosen as an arbitrary integer, this holds true for all integers.

Hence Proved: The function defined by

g(x)=x-[x]

is discontinuous at all integral points.

Q20EXERCISE 5.1

Is the function defined by $f(x)=x^{2}-\sin x+5$ continuous at $x=\pi$ ?

Solution

Given: The function is defined by

f(x)=x^{2}-\sin x+5

.

To Find: Whether the function

f(x)

is continuous at

x=\pi

.

Solution: We can analyze the continuity of

f(x)

by considering its component parts. Let's define three functions:

$h(x) = x^2$
$p(x) = \sin x$
$q(x) = 5$

Then, the given function can be written as

f(x) = h(x) - p(x) + q(x)

.

Now we check the continuity of each component function:

$h(x) = x^2$ is a polynomial function. Polynomial functions are continuous for all real numbers ( $x \in \mathbb{R}$ ). So, $h(x)$ is continuous everywhere.
$p(x) = \sin x$ is a trigonometric sine function. The sine function is continuous for all real numbers ( $x \in \mathbb{R}$ ). So, $p(x)$ is continuous everywhere.
$q(x) = 5$ is a constant function. Constant functions are continuous for all real numbers ( $x \in \mathbb{R}$ ). So, $q(x)$ is continuous everywhere.

According to the algebra of continuous functions, if two functions are continuous, their sum and difference are also continuous. Since

h(x)

,

p(x)

, and

q(x)

are all continuous functions on

\mathbb{R}

, their combination

f(x) = h(x) - p(x) + q(x)

is also continuous on

\mathbb{R}

.

Since the function

f(x)

is continuous for all real numbers, it must be continuous at any specific real number, including

x=\pi

.

Alternative Method (Using Limits): To check for continuity at

x=\pi

, we can compute the limit and the function's value at that point.

Limit at

x=\pi

:

\lim_{x \to \pi} f(x) = \lim_{x \to \pi} (x^2 - \sin x + 5)

= (\pi)^2 - \sin(\pi) + 5

Since

\sin(\pi) = 0

,

= \pi^2 - 0 + 5 = \pi^2 + 5

Value of the function at

x=\pi

:

f(\pi) = (\pi)^2 - \sin(\pi) + 5

= \pi^2 - 0 + 5 = \pi^2 + 5

Since

\lim_{x \to \pi} f(x) = f(\pi)

, the function is continuous at

x=\pi

.

Final Answer: Yes, the function defined by

f(x)=x^{2}-\sin x+5

is continuous at

x=\pi

.

Q21EXERCISE 5.1

Discuss the continuity of the following functions:

(a)

f(x)=\sin x+\cos x

(b)

f(x)=\sin x-\cos x

(c)

f(x)=\sin x \cdot \cos x

Solution

Given: The functions: (a)

f(x)=\sin x+\cos x

(b)

f(x)=\sin x-\cos x

(c)

f(x)=\sin x \cdot \cos x

To Discuss: The continuity of the given functions.

Solution: We know that the sine function,

g(x)=\sin x

, and the cosine function,

h(x)=\cos x

, are continuous for all real numbers

x \in \mathbb{R}

. We also know that if two functions are continuous, then their sum, difference, and product are also continuous.

(a) $f(x)=\sin x+\cos x$ Let

g(x) = \sin x

and

h(x) = \cos x

. Both are continuous functions on

\mathbb{R}

. The function

f(x)

is the sum of

g(x)

and

h(x)

. Since the sum of two continuous functions is continuous,

f(x) = \sin x + \cos x

is continuous for all

x \in \mathbb{R}

.

(b) $f(x)=\sin x-\cos x$ Let

g(x) = \sin x

and

h(x) = \cos x

. Both are continuous functions on

\mathbb{R}

. The function

f(x)

is the difference of

g(x)

and

h(x)

. Since the difference of two continuous functions is continuous,

f(x) = \sin x - \cos x

is continuous for all

x \in \mathbb{R}

.

(c) $f(x)=\sin x \cdot \cos x$ Let

g(x) = \sin x

and

h(x) = \cos x

. Both are continuous functions on

\mathbb{R}

. The function

f(x)

is the product of

g(x)

and

h(x)

. Since the product of two continuous functions is continuous,

f(x) = \sin x \cdot \cos x

is continuous for all

x \in \mathbb{R}

. Alternatively, we can write

f(x) = \frac{1}{2} (2 \sin x \cos x) = \frac{1}{2} \sin(2x)

. This is a composition of continuous functions (sine function and polynomial

2x

) and a product with a constant, which is continuous for all

x \in \mathbb{R}

.

Final Answer: All three functions, (a)

f(x)=\sin x+\cos x

, (b)

f(x)=\sin x-\cos x

, and (c)

f(x)=\sin x \cdot \cos x

, are continuous on the set of all real numbers

\mathbb{R}

.

Q22EXERCISE 5.1

Discuss the continuity of the cosine, cosecant, secant and cotangent functions.

Solution

To Discuss: The continuity of the cosine, cosecant, secant and cotangent functions.

Solution: We use the fact that

\sin x

and

\cos x

are continuous functions for all

x \in \mathbb{R}

. The quotient of two continuous functions

\frac{g(x)}{h(x)}

is continuous at all points where

h(x) \neq 0

.

1. Cosine function: The function is

f(x) = \cos x

. The cosine function is defined and continuous for all real numbers. Its graph is a smooth, unbroken curve. Domain of continuity:

x \in \mathbb{R}

.

2. Cosecant function: The function is

f(x) = \csc x = \frac{1}{\sin x}

. This function is a quotient of the constant function

1

(which is continuous) and the sine function

\sin x

(which is continuous). The function is continuous everywhere except at points where the denominator is zero.

\sin x = 0

when

x = n\pi

, for any integer

n \in \mathbb{Z}

. Domain of continuity:

f(x) = \csc x

is continuous for all real numbers

x

such that

x \neq n\pi

, where

n

is an integer.

3. Secant function: The function is

f(x) = \sec x = \frac{1}{\cos x}

. This function is a quotient of the constant function

1

and the cosine function

\cos x

. The function is continuous everywhere except at points where the denominator is zero.

\cos x = 0

when

x = (2n+1)\frac{\pi}{2}

, for any integer

n \in \mathbb{Z}

. Domain of continuity:

f(x) = \sec x

is continuous for all real numbers

x

such that

x \neq (2n+1)\frac{\pi}{2}

, where

n

is an integer.

4. Cotangent function: The function is

f(x) = \cot x = \frac{\cos x}{\sin x}

. This function is a quotient of the cosine function and the sine function, both of which are continuous. The function is continuous everywhere except at points where the denominator is zero.

\sin x = 0

when

x = n\pi

, for any integer

n \in \mathbb{Z}

. Domain of continuity:

f(x) = \cot x

is continuous for all real numbers

x

such that

x \neq n\pi

, where

n

is an integer.

Final Answer:

The cosine function is continuous for all $x \in \mathbb{R}$ .
The cosecant function is continuous for all $x \in \mathbb{R} - \{n\pi : n \in \mathbb{Z}\}$ .
The secant function is continuous for all $x \in \mathbb{R} - \{(2n+1)\frac{\pi}{2} : n \in \mathbb{Z}\}$ .
The cotangent function is continuous for all $x \in \mathbb{R} - \{n\pi : n \in \mathbb{Z}\}$ .

Q23EXERCISE 5.1

Find all points of discontinuity of $f$ , where $f(x)= \begin{cases}\frac{\sin x}{x}, & \text { if } x<0 \ x+1, & \text { if } x \geq 0\end{cases}$

Solution

Given: The function

f

is defined as:

f(x)= \begin{cases}\frac{\sin x}{x}, & \text { if } x<0 \ x+1, & \text { if } x \geq 0\end{cases}

To Find: All points of discontinuity of

f

.

Solution: We analyze the continuity of the function in three parts: for

x < 0

, for

x > 0

, and at

x = 0

.

Case 1: For $x < 0$ In this interval,

f(x) = \frac{\sin x}{x}

. The function

\sin x

is continuous for all real numbers. The function

x

(identity function) is also continuous for all real numbers. The quotient of two continuous functions is continuous, provided the denominator is not zero. For

x < 0

, the denominator

x

is not zero. Therefore,

f(x)

is continuous for all

x < 0

.

Case 2: For $x > 0$ In this interval,

f(x) = x+1

. This is a polynomial function, and polynomial functions are continuous for all real numbers. Therefore,

f(x)

is continuous for all

x > 0

.

Case 3: At $x = 0$ A function is continuous at a point

c

if

\lim_{x \to c} f(x) = f(c)

. First, we find the value of the function at

x=0

. From the definition, for

x \geq 0

,

f(x) = x+1

. So,

f(0) = 0+1 = 1

Next, we find the limit of the function as

x

approaches 0. We need to evaluate the left-hand limit (LHL) and the right-hand limit (RHL).

Left-Hand Limit (LHL): For

x < 0

,

f(x) = \frac{\sin x}{x}

.

\text{LHL} = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\sin x}{x}

Using the standard trigonometric limit, we have:

\text{LHL} = 1

Right-Hand Limit (RHL): For

x > 0

,

f(x) = x+1

.

\text{RHL} = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x+1) = 0+1 = 1

Since LHL = RHL = 1, the limit exists and is equal to 1.

\lim_{x \to 0} f(x) = 1

Now we compare the limit with the function's value at

x=0

.

\lim_{x \to 0} f(x) = 1 \quad \text{and} \quad f(0) = 1

Since

\lim_{x \to 0} f(x) = f(0)

, the function

f

is continuous at

x=0

.

Conclusion: The function

f(x)

is continuous for

x<0

,

x>0

, and at

x=0

. Therefore, the function is continuous for all real numbers.

Final Answer: There are no points of discontinuity for the given function

f(x)

.

Q24EXERCISE 5.1

Determine if $f$ defined by $f(x)= \begin{cases}x^{2} \sin \frac{1}{x}, & \text { if } x \neq 0 \ 0, & \text { if } x=0\end{cases}$ is a continuous function?

Solution

Given: The function

f

is defined as:

f(x)= \begin{cases}x^{2} \sin \frac{1}{x}, & \text { if } x \neq 0 \ 0, & \text { if } x=0\end{cases}

To Determine: If

f

is a continuous function.

Solution: We check for continuity for

x \neq 0

and at

x=0

.

Case 1: For $x \neq 0$ When

x \neq 0

,

f(x) = x^2 \sin \frac{1}{x}

. The function

g(x) = x^2

is a polynomial and is continuous for all

x \in \mathbb{R}

. The function

h(x) = \frac{1}{x}

is a rational function and is continuous for all

x \neq 0

. The function

k(t) = \sin t

is continuous for all

t \in \mathbb{R}

. The composition

k(h(x)) = \sin(\frac{1}{x})

is continuous for all

x \neq 0

. Since

f(x)

is the product of two continuous functions for

x \neq 0

(i.e.,

x^2

and

\sin(\frac{1}{x})

),

f(x)

is continuous for all

x \neq 0

.

Case 2: At $x=0$ For

f

to be continuous at

x=0

, we must have

\lim_{x \to 0} f(x) = f(0)

. From the definition of the function, we are given

f(0) = 0

. Now, we need to evaluate the limit:

\lim_{x \to 0} f(x) = \lim_{x \to 0} x^2 \sin \frac{1}{x}

We know that for any real number

t

, the value of

\sin t

lies in the interval

[-1, 1]

. So, for any

x \neq 0

, we have:

-1 \leq \sin \frac{1}{x} \leq 1

Multiplying the inequality by

x^2

(which is always non-negative), we get:

-x^2 \leq x^2 \sin \frac{1}{x} \leq x^2

Now, we take the limit as

x \to 0

for all parts of the inequality.

\lim_{x \to 0} (-x^2) = 0

\lim_{x \to 0} (x^2) = 0

By the Squeeze Theorem (or Sandwich Theorem), since

x^2 \sin \frac{1}{x}

is squeezed between two functions that both approach 0 as

x \to 0

, we have:

\lim_{x \to 0} x^2 \sin \frac{1}{x} = 0

So,

\lim_{x \to 0} f(x) = 0

.

Comparing the limit with the function's value at

x=0

:

\lim_{x \to 0} f(x) = 0 \quad \text{and} \quad f(0) = 0

Since

\lim_{x \to 0} f(x) = f(0)

, the function is continuous at

x=0

.

Conclusion: The function is continuous for all

x \neq 0

and is also continuous at

x=0

. Therefore, the function

f

is continuous for all real numbers.

Final Answer: Yes, the function

f(x)

is a continuous function.

Q25EXERCISE 5.1

Examine the continuity of $f$ , where $f$ is defined by $f(x)= \begin{cases}\sin x-\cos x, & \text { if } x \neq 0 \ -1, & \text { if } x=0\end{cases}$

Solution

Given: The function

f

is defined as:

f(x)= \begin{cases}\sin x-\cos x, & \text { if } x \neq 0 \ -1, & \text { if } x=0\end{cases}

To Find: Examine the continuity of

f

.

Solution: We need to check the continuity of the function for

x \neq 0

and at

x=0

.

Case 1: For $x \neq 0$ In this case,

f(x) = \sin x - \cos x

. We know that

\sin x

and

\cos x

are continuous functions for all real numbers. The difference of two continuous functions is also a continuous function. Therefore,

f(x) = \sin x - \cos x

is continuous for all

x \neq 0

.

Case 2: At $x = 0$ For the function to be continuous at

x=0

, we must have

\lim_{x \to 0} f(x) = f(0)

. From the definition of the function, we are given:

f(0) = -1

Now, we evaluate the limit of the function as

x

approaches 0. For

x \to 0

, we use the definition

f(x) = \sin x - \cos x

.

\lim_{x \to 0} f(x) = \lim_{x \to 0} (\sin x - \cos x)

Since

\sin x

and

\cos x

are continuous at

x=0

, we can substitute

x=0

directly into the expression:

\lim_{x \to 0} (\sin x - \cos x) = \sin(0) - \cos(0) = 0 - 1 = -1

So, we have:

\lim_{x \to 0} f(x) = -1

Now, we compare the limit with the function's value at

x=0

.

\lim_{x \to 0} f(x) = -1 \quad \text{and} \quad f(0) = -1

Since

\lim_{x \to 0} f(x) = f(0)

, the function

f

is continuous at

x=0

.

Conclusion: The function

f(x)

is continuous for all

x \neq 0

and is also continuous at

x=0

. Thus, the function is continuous for all real numbers.

Final Answer: The function

f

is a continuous function for all

x \in \mathbb{R}

.

Q26EXERCISE 5.1

Find the values of $k$ so that the function $f$ is continuous at the indicated point in Exercises 26 to 29. 26. $f(x)=\left\{\begin{array}{ll}\frac{k \cos x}{\pi-2 x}, & \text { if } x \neq \frac{\pi}{2} \ 3, & \text { if } x=\frac{\pi}{2}\end{array}\right.$ \quad at $x=\frac{\pi}{2}$

Solution

Given: The function

f(x)

is defined as:

f(x)=\begin{cases}\frac{k \cos x}{\pi-2 x}, & \text { if } x \neq \frac{\pi}{2} \ 3, & \text { if } x=\frac{\pi}{2}\end{cases}

The function is continuous at

x=\frac{\pi}{2}

.

To Find: The value of

k

.

Solution: For a function to be continuous at a point

x=c

, we must have:

\lim_{x \to c} f(x) = f(c)

Here,

c = \frac{\pi}{2}

. We are given

f(\frac{\pi}{2}) = 3

.

For continuity at

x = \frac{\pi}{2}

, we need:

\lim_{x \to \frac{\pi}{2}} f(x) = f\left(\frac{\pi}{2}\right)

\lim_{x \to \frac{\pi}{2}} \frac{k \cos x}{\pi-2 x} = 3

To evaluate the limit, let's make a substitution. Let

x = \frac{\pi}{2} + h

. As

x \to \frac{\pi}{2}

,

h \to 0

. Also,

\pi - 2x = \pi - 2(\frac{\pi}{2} + h) = \pi - \pi - 2h = -2h

. And

\cos x = \cos(\frac{\pi}{2} + h) = -\sin h

.

Substituting these into the limit expression:

\lim_{h \to 0} \frac{k(-\sin h)}{-2h} = \lim_{h \to 0} \frac{k}{2} \left(\frac{\sin h}{h}\right)

We know that

\lim_{h \to 0} \frac{\sin h}{h} = 1

. So, the limit is:

\frac{k}{2} \times 1 = \frac{k}{2}

Equating this to the value of the function at the point:

\frac{k}{2} = 3

k = 6

Final Answer: The value of

k

is 6.

Q27EXERCISE 5.1

Find the values of $k$ so that the function $f$ is continuous at the indicated point in Exercises 26 to 29. 27. $f(x)=\left\{\begin{array}{ll}k x^{2}, & \text { if } x \leq 2 \ 3, & \text { if } x>2\end{array}\right.$ \quad at $x=2$

Solution

Given: The function

f(x)

is defined as:

f(x)=\begin{cases}k x^{2}, & \text { if } x \leq 2 \ 3, & \text { if } x>2\end{cases}

The function is continuous at

x=2

.

To Find: The value of

k

.

Solution: For a function to be continuous at a point

x=c

, the left-hand limit (LHL) must equal the right-hand limit (RHL), and this common value must be equal to the function's value at that point. That is:

\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)

Here,

c=2

.

First, let's find the value of the function at

x=2

: For

x \leq 2

,

f(x) = kx^2

. So,

f(2) = k(2)^2 = 4k

.

Next, we find the Left-Hand Limit (LHL) at

x=2

:

\text{LHL} = \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (kx^2) = k(2)^2 = 4k

Now, we find the Right-Hand Limit (RHL) at

x=2

:

\text{RHL} = \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (3) = 3

For the function to be continuous at

x=2

, we must have LHL = RHL.

4k = 3

k = \frac{3}{4}

Final Answer: The value of

k

is

\frac{3}{4}

.

Q28EXERCISE 5.1

Find the values of $k$ so that the function $f$ is continuous at the indicated point in Exercises 26 to 29. 28. $f(x)=\left\{\begin{array}{ll}k x+1, & \text { if } x \leq \pi \ \cos x, & \text { if } x>\pi\end{array}\right.$ \quad at $x=\pi$

Solution

Given: The function

f(x)

is defined as:

f(x)=\begin{cases}k x+1, & \text { if } x \leq \pi \ \cos x, & \text { if } x>\pi\end{cases}

The function is continuous at

x=\pi

.

To Find: The value of

k

.

Solution: For a function to be continuous at a point

x=c

, the left-hand limit (LHL) must equal the right-hand limit (RHL), and this common value must be equal to the function's value at that point. That is:

\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)

Here,

c=\pi

.

First, let's find the value of the function at

x=\pi

: For

x \leq \pi

,

f(x) = kx+1

. So,

f(\pi) = k\pi + 1

.

Next, we find the Left-Hand Limit (LHL) at

x=\pi

:

\text{LHL} = \lim_{x \to \pi^-} f(x) = \lim_{x \to \pi^-} (kx+1) = k\pi + 1

Now, we find the Right-Hand Limit (RHL) at

x=\pi

:

\text{RHL} = \lim_{x \to \pi^+} f(x) = \lim_{x \to \pi^+} (\cos x) = \cos(\pi) = -1

For the function to be continuous at

x=\pi

, we must have LHL = RHL.

k\pi + 1 = -1

k\pi = -2

k = -\frac{2}{\pi}

Final Answer: The value of

k

is

-\frac{2}{\pi}

.

Q29EXERCISE 5.1

Find the values of $k$ so that the function $f$ is continuous at the indicated point in Exercises 26 to 29. 29. $f(x)=\left\{\begin{array}{ll}k x+1, & \text { if } x \leq 5 \ 3 x-5, & \text { if } x>5\end{array}\right.$ \quad at $x=5$

Solution

Given: The function

f(x)

is defined as:

f(x)=\begin{cases}k x+1, & \text { if } x \leq 5 \ 3x-5, & \text { if } x>5\end{cases}

The function is continuous at

x=5

.

To Find: The value of

k

.

Solution: For a function to be continuous at a point

x=c

, the left-hand limit (LHL) must equal the right-hand limit (RHL), and this common value must be equal to the function's value at that point. That is:

\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)

Here,

c=5

.

First, let's find the value of the function at

x=5

: For

x \leq 5

,

f(x) = kx+1

. So,

f(5) = k(5) + 1 = 5k+1

.

Next, we find the Left-Hand Limit (LHL) at

x=5

:

\text{LHL} = \lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} (kx+1) = k(5) + 1 = 5k+1

Now, we find the Right-Hand Limit (RHL) at

x=5

:

\text{RHL} = \lim_{x \to 5^+} f(x) = \lim_{x \to 5^+} (3x-5) = 3(5) - 5 = 15 - 5 = 10

For the function to be continuous at

x=5

, we must have LHL = RHL.

5k + 1 = 10

5k = 9

k = \frac{9}{5}

Final Answer: The value of

k

is

\frac{9}{5}

.

Q30EXERCISE 5.1

Find the values of $a$ and $b$ such that the function defined by $f(x)= \begin{cases}5, & \text { if } x \leq 2 \ a x+b, & \text { if } 2<x<10 \ 21, & \text { if } x \geq 10\end{cases}$ is a continuous function.

Solution

Given: The function

f(x)

is defined as:

f(x)= \begin{cases}5, & \text { if } x \leq 2 \ a x+b, & \text { if } 2<x<10 \ 21, & \text { if } x \geq 10\end{cases}

The function

f(x)

is a continuous function.

To Find: The values of

a

and

b

.

Solution: The function is defined in three pieces. The individual functions

f(x)=5

,

f(x)=ax+b

(a polynomial), and

f(x)=21

are continuous in their respective domains. For

f(x)

to be a continuous function for all real numbers, it must be continuous at the points where the definition changes, which are

x=2

and

x=10

.

Continuity at x = 2: For

f

to be continuous at

x=2

, we must have

\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2)

.

Left-Hand Limit (LHL):

\text{LHL} = \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (5) = 5

Right-Hand Limit (RHL):

\text{RHL} = \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (ax+b) = a(2) + b = 2a+b

Since the function is continuous at

x=2

, LHL = RHL.

2a+b = 5 \quad \dots(1)

Continuity at x = 10: For

f

to be continuous at

x=10

, we must have

\lim_{x \to 10^-} f(x) = \lim_{x \to 10^+} f(x) = f(10)

.

Left-Hand Limit (LHL):

\text{LHL} = \lim_{x \to 10^-} f(x) = \lim_{x \to 10^-} (ax+b) = a(10) + b = 10a+b

Right-Hand Limit (RHL):

\text{RHL} = \lim_{x \to 10^+} f(x) = \lim_{x \to 10^+} (21) = 21

Since the function is continuous at

x=10

, LHL = RHL.

10a+b = 21 \quad \dots(2)

Now we have a system of two linear equations with two variables,

a

and

b

:

$2a + b = 5$
$10a + b = 21$

Subtracting equation (1) from equation (2):

(10a + b) - (2a + b) = 21 - 5

8a = 16

a = 2

Substitute the value of

a=2

into equation (1):

2(2) + b = 5

4 + b = 5

b = 1

Thus, for the function to be continuous, the values of

a

and

b

must be 2 and 1, respectively.

Final Answer: The values are

a=2

and

b=1

.

Q31EXERCISE 5.1

Show that the function defined by $f(x)=\cos \left(x^{2}\right)$ is a continuous function.

Solution

To Show: The function defined by

f(x)=\cos(x^2)

is a continuous function.

Proof: The given function is

f(x) = \cos(x^2)

. This function can be expressed as the composition of two functions. Let

g(x) = x^2

and

h(x) = \cos x

. Then,

(h \circ g)(x) = h(g(x)) = h(x^2) = \cos(x^2) = f(x)

.

We know that:

The function $g(x) = x^2$ is a polynomial function. Since every polynomial function is continuous for all real numbers, $g(x)$ is continuous on $\mathbb{R}$ .
The function $h(x) = \cos x$ is a cosine function. The cosine function is continuous for all real numbers, so $h(x)$ is continuous on $\mathbb{R}$ .

According to the theorem on the composition of continuous functions, if two functions

g

and

h

are continuous, then their composition

(h \circ g)

is also continuous.

Since both

g(x)

and

h(x)

are continuous functions, their composition

f(x) = (h \circ g)(x) = \cos(x^2)

is also a continuous function for all

x \in \mathbb{R}

.

Hence Proved.

Q32EXERCISE 5.1

Show that the function defined by $f(x)=|\cos x|$ is a continuous function.

Solution

To Show: The function defined by

f(x)=|\cos x|

is a continuous function.

Proof: The given function is

f(x) = |\cos x|

. This function can be expressed as the composition of two functions. Let

g(x) = \cos x

and

h(x) = |x|

. Then,

(h \circ g)(x) = h(g(x)) = h(\cos x) = |\cos x| = f(x)

.

We know that:

The function $g(x) = \cos x$ is a cosine function. The cosine function is continuous for all real numbers, so $g(x)$ is continuous on $\mathbb{R}$ .
The function $h(x) = |x|$ is the modulus function. The modulus function is continuous for all real numbers, so $h(x)$ is continuous on $\mathbb{R}$ .

According to the theorem on the composition of continuous functions, if two functions

g

and

h

are continuous, then their composition

(h \circ g)

is also continuous.

Since both

g(x)

and

h(x)

are continuous functions, their composition

f(x) = (h \circ g)(x) = |\cos x|

is also a continuous function for all

x \in \mathbb{R}

.

Hence Proved.

Q33EXERCISE 5.1

Examine that $\sin |x|$ is a continuous function.

Solution

To Examine: Whether

\sin |x|

is a continuous function.

Solution: Let the given function be

f(x) = \sin |x|

. This function can be expressed as the composition of two functions. Let

g(x) = |x|

and

h(x) = \sin x

. Then,

(h \circ g)(x) = h(g(x)) = h(|x|) = \sin |x| = f(x)

.

We know that:

The function $g(x) = |x|$ is the modulus function. The modulus function is continuous for all real numbers, so $g(x)$ is continuous on $\mathbb{R}$ .
The function $h(x) = \sin x$ is a sine function. The sine function is continuous for all real numbers, so $h(x)$ is continuous on $\mathbb{R}$ .

According to the theorem on the composition of continuous functions, if two functions

g

and

h

are continuous, then their composition

(h \circ g)

is also continuous.

Since both

g(x)

and

h(x)

are continuous functions, their composition

f(x) = (h \circ g)(x) = \sin |x|

is also a continuous function for all

x \in \mathbb{R}

.

Final Answer: The function

\sin |x|

is a continuous function.

Q34EXERCISE 5.1

Find all the points of discontinuity of $f$ defined by $f(x)=|x|-|x+1|$ .

Solution

Given: The function

f(x) = |x| - |x+1|

.

To Find: All the points of discontinuity of

f

.

Solution: The function

f(x)

is defined as the difference of two functions. Let's define them as:

g(x) = |x|

h(x) = |x+1|

So,

f(x) = g(x) - h(x)

.

Continuity of $g(x)$ : The function $g(x) = |x|$ is the modulus function. We know that the modulus function is continuous for all real numbers. So, $g(x)$ is continuous on $\mathbb{R}$ .
Continuity of $h(x)$ : The function $h(x) = |x+1|$ can be seen as a composition of two functions. Let $p(x) = x+1$ and $q(x) = |x|$ . Then $h(x) = q(p(x)) = |x+1|$ . The function $p(x) = x+1$ is a polynomial function, which is continuous everywhere. The function $q(x) = |x|$ is the modulus function, which is also continuous everywhere. Since the composition of two continuous functions is continuous, $h(x) = |x+1|$ is continuous for all $x \in \mathbb{R}$ .
Continuity of $f(x)$ : According to the algebra of continuous functions, if two functions $g(x)$ and $h(x)$ are continuous on a domain, then their difference, $f(x) = g(x) - h(x)$ , is also continuous on that domain.

Since both

g(x) = |x|

and

h(x) = |x+1|

are continuous for all real numbers, their difference

f(x) = |x| - |x+1|

is also continuous for all real numbers.

A function that is continuous everywhere has no points of discontinuity.

Final Answer: There are no points of discontinuity for the function

f(x) = |x| - |x+1|

.

Q1EXERCISE 5.2

Differentiate the functions with respect to $x$ in Exercises 1 to 8 . $\sin \left(x^{2}+5\right)$

Solution

Given: The function

f(x) = \sin(x^2 + 5)

.

To Find: The derivative of the function with respect to

x

.

Solution: We need to find

\frac{d}{dx} \sin(x^2 + 5)

. We will use the chain rule for differentiation, which states that if

y = f(g(x))

, then

\frac{dy}{dx} = f'(g(x)) \cdot g'(x)

.

Let's identify the outer function and the inner function. Outer function:

f(u) = \sin u

Inner function:

g(x) = x^2 + 5

First, find the derivatives of the outer and inner functions: Derivative of the outer function with respect to its argument:

f'(u) = \frac{d}{du}(\sin u) = \cos u

. Derivative of the inner function with respect to

x

:

g'(x) = \frac{d}{dx}(x^2 + 5) = 2x + 0 = 2x

.

Now, apply the chain rule:

\frac{d}{dx} \sin(x^2 + 5) = f'(g(x)) \cdot g'(x)

= \cos(x^2 + 5) \cdot (2x)

= 2x \cos(x^2 + 5)

Final Answer: The derivative of

\sin(x^2 + 5)

with respect to

x

is

2x \cos(x^2 + 5)

.

Q2EXERCISE 5.2

Differentiate the functions with respect to $x$ in Exercises 1 to 8 . 2. $\cos (\sin x)$

Solution

Given function:

f(x) = \cos(\sin x)

To Find: The derivative of the function with respect to

x

.

Solution: Let

y = \cos(\sin x)

. We will differentiate

y

with respect to

x

using the chain rule. Let

u = \sin x

. Then

y = \cos u

. According to the chain rule, we have:

\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}

First, we find the derivative of

y

with respect to

u

:

\frac{dy}{du} = \frac{d}{du}(\cos u) = -\sin u

Substituting

u = \sin x

back, we get:

\frac{dy}{du} = -\sin(\sin x)

Next, we find the derivative of

u

with respect to

x

:

\frac{du}{dx} = \frac{d}{dx}(\sin x) = \cos x

Now, we multiply these results to get

\frac{dy}{dx}

:

\frac{dy}{dx} = -\sin(\sin x) \times \cos x

\frac{dy}{dx} = -\cos x \sin(\sin x)

Final Answer: The derivative of

\cos(\sin x)

with respect to

x

is

-\cos x \sin(\sin x)

.

Q3EXERCISE 5.2

$\sin (a x+b)$

Solution

Given function:

f(x) = \sin(ax+b)

To Find: The derivative of the function with respect to

x

.

Solution: Let

y = \sin(ax+b)

. We will differentiate

y

with respect to

x

using the chain rule. Let

u = ax+b

. Then

y = \sin u

. According to the chain rule, we have:

\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}

First, we find the derivative of

y

with respect to

u

:

\frac{dy}{du} = \frac{d}{du}(\sin u) = \cos u

Substituting

u = ax+b

back, we get:

\frac{dy}{du} = \cos(ax+b)

Next, we find the derivative of

u

with respect to

x

:

\frac{du}{dx} = \frac{d}{dx}(ax+b) = \frac{d}{dx}(ax) + \frac{d}{dx}(b) = a(1) + 0 = a

Now, we multiply these results to get

\frac{dy}{dx}

:

\frac{dy}{dx} = \cos(ax+b) \times a

\frac{dy}{dx} = a \cos(ax+b)

Final Answer: The derivative of

\sin(ax+b)

with respect to

x

is

a \cos(ax+b)

.

Q4EXERCISE 5.2

$\sec (\tan (\sqrt{x}))$

Solution

Given function:

f(x) = \sec(\tan(\sqrt{x}))

To Find: The derivative of the function with respect to

x

.

Solution: Let

y = \sec(\tan(\sqrt{x}))

. We will differentiate

y

with respect to

x

by applying the chain rule multiple times.

\frac{dy}{dx} = \frac{d}{dx} [\sec(\tan(\sqrt{x}))]

Using the chain rule,

\frac{d}{dx}(\sec(u)) = \sec(u)\tan(u) \frac{du}{dx}

. Here,

u = \tan(\sqrt{x})

.

\frac{dy}{dx} = \sec(\tan(\sqrt{x})) \tan(\tan(\sqrt{x})) \times \frac{d}{dx}(\tan(\sqrt{x}))

Now, we need to find the derivative of

\tan(\sqrt{x})

. We apply the chain rule again.

\frac{d}{dx}(\tan(v)) = \sec^2(v) \frac{dv}{dx}

. Here,

v = \sqrt{x}

.

\frac{d}{dx}(\tan(\sqrt{x})) = \sec^2(\sqrt{x}) \times \frac{d}{dx}(\sqrt{x})

Finally, we find the derivative of

\sqrt{x} = x^{1/2}

.

\frac{d}{dx}(\sqrt{x}) = \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{\frac{1}{2}-1} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}

Now, substitute the derivatives back into the expression for

\frac{dy}{dx}

:

\frac{dy}{dx} = \sec(\tan(\sqrt{x})) \tan(\tan(\sqrt{x})) \times [\sec^2(\sqrt{x}) \times \frac{1}{2\sqrt{x}}]

Rearranging the terms for the final answer:

\frac{dy}{dx} = \frac{\sec(\tan(\sqrt{x})) \tan(\tan(\sqrt{x})) \sec^2(\sqrt{x})}{2\sqrt{x}}

Final Answer: The derivative of

\sec(\tan(\sqrt{x}))

with respect to

x

is

\frac{\sec(\tan(\sqrt{x})) \tan(\tan(\sqrt{x})) \sec^2(\sqrt{x})}{2\sqrt{x}}

.

Q5EXERCISE 5.2

$\frac{\sin (a x+b)}{\cos (c x+d)}$

Solution

Given function:

f(x) = \frac{\sin(ax+b)}{\cos(cx+d)}

To Find: The derivative of the function with respect to

x

.

Solution: Let

y = \frac{\sin(ax+b)}{\cos(cx+d)}

. This is a quotient of two functions. Let

u = \sin(ax+b)

and

v = \cos(cx+d)

. We use the quotient rule for differentiation:

\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}

First, we find the derivatives of

u

and

v

with respect to

x

using the chain rule. For

u = \sin(ax+b)

:

\frac{du}{dx} = \frac{d}{dx}(\sin(ax+b)) = \cos(ax+b) \cdot \frac{d}{dx}(ax+b) = a\cos(ax+b)

For

v = \cos(cx+d)

:

\frac{dv}{dx} = \frac{d}{dx}(\cos(cx+d)) = -\sin(cx+d) \cdot \frac{d}{dx}(cx+d) = -c\sin(cx+d)

Now, substitute these derivatives into the quotient rule formula:

\frac{dy}{dx} = \frac{\cos(cx+d) \cdot [a\cos(ax+b)] - \sin(ax+b) \cdot [-c\sin(cx+d)]}{[\cos(cx+d)]^2}

Simplify the numerator:

\frac{dy}{dx} = \frac{a\cos(ax+b)\cos(cx+d) + c\sin(ax+b)\sin(cx+d)}{\cos^2(cx+d)}

We can separate this into two fractions for a more simplified form:

\frac{dy}{dx} = \frac{a\cos(ax+b)\cos(cx+d)}{\cos^2(cx+d)} + \frac{c\sin(ax+b)\sin(cx+d)}{\cos^2(cx+d)}

\frac{dy}{dx} = a\cos(ax+b) \cdot \frac{1}{\cos(cx+d)} + c\sin(ax+b) \cdot \frac{\sin(cx+d)}{\cos(cx+d)} \cdot \frac{1}{\cos(cx+d)}

Using the identities

\sec\theta = \frac{1}{\cos\theta}

and

\tan\theta = \frac{\sin\theta}{\cos\theta}

:

\frac{dy}{dx} = a\cos(ax+b)\sec(cx+d) + c\sin(ax+b)\tan(cx+d)\sec(cx+d)

Final Answer: The derivative of

\frac{\sin(ax+b)}{\cos(cx+d)}

is

a\cos(ax+b)\sec(cx+d) + c\sin(ax+b)\tan(cx+d)\sec(cx+d)

.

Q6EXERCISE 5.2

$\cos x^{3} \cdot \sin ^{2}\left(x^{5}\right)$

Solution

Given function:

f(x) = \cos x^3 \cdot \sin^2(x^5)

To Find: The derivative of the function with respect to

x

.

Solution: Let

y = \cos(x^3) \cdot \sin^2(x^5)

. This is a product of two functions. Let

u = \cos(x^3)

and

v = \sin^2(x^5) = (\sin(x^5))^2

. We use the product rule for differentiation:

\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}

First, we find the derivatives of

u

and

v

with respect to

x

using the chain rule. For

u = \cos(x^3)

:

\frac{du}{dx} = \frac{d}{dx}(\cos(x^3)) = -\sin(x^3) \cdot \frac{d}{dx}(x^3) = -\sin(x^3) \cdot 3x^2 = -3x^2\sin(x^3)

For

v = (\sin(x^5))^2

:

\frac{dv}{dx} = \frac{d}{dx}[(\sin(x^5))^2]

Using the power rule and chain rule:

\frac{dv}{dx} = 2(\sin(x^5))^{2-1} \cdot \frac{d}{dx}(\sin(x^5))

\frac{dv}{dx} = 2\sin(x^5) \cdot [\cos(x^5) \cdot \frac{d}{dx}(x^5)]

\frac{dv}{dx} = 2\sin(x^5)\cos(x^5) \cdot 5x^4

Using the identity

2\sin A \cos A = \sin(2A)

, we can simplify this:

\frac{dv}{dx} = 5x^4 (2\sin(x^5)\cos(x^5)) = 5x^4\sin(2x^5)

Now, substitute these derivatives into the product rule formula:

\frac{dy}{dx} = (\cos(x^3)) \cdot [5x^4\sin(2x^5)] + (\sin^2(x^5)) \cdot [-3x^2\sin(x^3)]

Simplify and rearrange the terms:

\frac{dy}{dx} = 5x^4\cos(x^3)\sin(2x^5) - 3x^2\sin^2(x^5)\sin(x^3)

Final Answer: The derivative of

\cos x^3 \cdot \sin^2(x^5)

with respect to

x

is

5x^4\cos(x^3)\sin(2x^5) - 3x^2\sin^2(x^5)\sin(x^3)

.

Q7EXERCISE 5.2

$2 \sqrt{\cot \left(x^{2}\right)}$

Solution

Given: The function

y = 2 \sqrt{\cot(x^2)}

.

To Find: The derivative of the function,

\frac{dy}{dx}

.

Solution: We need to differentiate the function

y = 2 \sqrt{\cot(x^2)}

with respect to

x

. We will use the chain rule. Let the function be written as

y = 2(\cot(x^2))^{1/2}

.

Differentiating

y

with respect to

x

:

\frac{dy}{dx} = \frac{d}{dx} [2(\cot(x^2))^{1/2}]

Using the chain rule,

\frac{d}{dx}(f(g(h(x)))) = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x)

. Here, the outer function is

2u^{1/2}

, the middle function is

\cot(v)

, and the inner function is

x^2

.

\frac{dy}{dx} = 2 \cdot \frac{1}{2}(\cot(x^2))^{\frac{1}{2}-1} \cdot \frac{d}{dx}(\cot(x^2))

= (\cot(x^2))^{-1/2} \cdot \frac{d}{dx}(\cot(x^2))

= \frac{1}{\sqrt{\cot(x^2)}} \cdot \frac{d}{dx}(\cot(x^2))

Now, we differentiate

\cot(x^2)

using the chain rule again:

\frac{d}{dx}(\cot(x^2)) = -\csc^2(x^2) \cdot \frac{d}{dx}(x^2)

= -\csc^2(x^2) \cdot (2x)

Substitute this back into the expression for

\frac{dy}{dx}

:

\frac{dy}{dx} = \frac{1}{\sqrt{\cot(x^2)}} \cdot (-2x \csc^2(x^2))

\frac{dy}{dx} = \frac{-2x \csc^2(x^2)}{\sqrt{\cot(x^2)}}

We can also express

\csc^2(x^2)

as

\frac{1}{\sin^2(x^2)}

and

\cot(x^2)

as

\frac{\cos(x^2)}{\sin(x^2)}

to simplify further if needed, but the above form is a complete answer.

Final Answer: The derivative of

2 \sqrt{\cot(x^2)}

is

\frac{-2x \csc^2(x^2)}{\sqrt{\cot(x^2)}}

.

Q8EXERCISE 5.2

$\cos (\sqrt{x})$

Solution

Given: The function

y = \cos(\sqrt{x})

.

To Find: The derivative of the function,

\frac{dy}{dx}

.

Solution: We need to differentiate the function

y = \cos(\sqrt{x})

with respect to

x

. We will use the chain rule.

Let

u = \sqrt{x}

. Then

y = \cos(u)

. By the chain rule,

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

.

First, find

\frac{dy}{du}

:

y = \cos(u)

\frac{dy}{du} = -\sin(u)

Next, find

\frac{du}{dx}

:

u = \sqrt{x} = x^{1/2}

\frac{du}{dx} = \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{\frac{1}{2}-1} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}

Now, combine these results:

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = -\sin(u) \cdot \frac{1}{2\sqrt{x}}

Substitute

u = \sqrt{x}

back into the expression:

\frac{dy}{dx} = -\sin(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}}

\frac{dy}{dx} = -\frac{\sin(\sqrt{x})}{2\sqrt{x}}

Final Answer: The derivative of

\cos(\sqrt{x})

is

-\frac{\sin(\sqrt{x})}{2\sqrt{x}}

.

Q9EXERCISE 5.2

Prove that the function $f$ given by $f(x)=|x-1|, x \in \mathbf{R}$ is not differentiable at $x=1$ .

Solution

To Prove: The function

f

given by

f(x)=|x-1|, x \in \mathbf{R}

is not differentiable at

x=1

.

Proof: A function

f(x)

is differentiable at a point

x=c

if the Left-Hand Derivative (LHD) and the Right-Hand Derivative (RHD) at

x=c

exist and are equal.

The derivative of a function

f

at a point

c

is given by

f'(c) = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h}

.

Here, we need to check the differentiability at

x=1

. So,

c=1

. First, we find the value of the function at

x=1

:

f(1) = |1-1| = |0| = 0

.

1. Calculate the Left-Hand Derivative (LHD) at $x=1$ : LHD

= \lim_{h \to 0^-} \frac{f(1+h) - f(1)}{h}

= \lim_{h \to 0^-} \frac{|(1+h)-1| - 0}{h}

= \lim_{h \to 0^-} \frac{|h|}{h}

Since

h \to 0^-

,

h

is a small negative number, so

h < 0

. Therefore,

|h| = -h

.

LHD = \lim_{h \to 0^-} \frac{-h}{h} = \lim_{h \to 0^-} (-1) = -1

2. Calculate the Right-Hand Derivative (RHD) at $x=1$ : RHD

= \lim_{h \to 0^+} \frac{f(1+h) - f(1)}{h}

= \lim_{h \to 0^+} \frac{|(1+h)-1| - 0}{h}

= \lim_{h \to 0^+} \frac{|h|}{h}

Since

h \to 0^+

,

h

is a small positive number, so

h > 0

. Therefore,

|h| = h

.

RHD = \lim_{h \to 0^+} \frac{h}{h} = \lim_{h \to 0^+} (1) = 1

Conclusion: Since the Left-Hand Derivative at

x=1

is

-1

and the Right-Hand Derivative at

x=1

is

1

, we have:

LHD \neq RHD

Because the left-hand and right-hand derivatives are not equal, the function

f(x) = |x-1|

is not differentiable at

x=1

.

Hence Proved.

Q10EXERCISE 5.2

Prove that the greatest integer function defined by $f(x)=[x], 0<x<3$ is not differentiable at $x=1$ and $x=2$ .

Solution

To Prove: The greatest integer function defined by

f(x)=[x]

, for

0 < x < 3

, is not differentiable at

x=1

and

x=2

.

Proof: We will check the differentiability at

x=1

and

x=2

separately by comparing the Left-Hand Derivative (LHD) and Right-Hand Derivative (RHD) at each point.

Case 1: Differentiability at $x=1$ First, we find the value of the function at

x=1

:

f(1) = [1] = 1

.

Left-Hand Derivative (LHD) at $x=1$ : LHD $= \lim_{h \to 0^-} \frac{f(1+h) - f(1)}{h}$ $= \lim_{h \to 0^-} \frac{[1+h] - 1}{h}$ As $h \to 0^-$ , $h$ is a small negative number. This means $0 < 1+h < 1$ . For example, if $h=-0.1$ , $1+h=0.9$ . The greatest integer less than or equal to $1+h$ is $0$ . So, $[1+h] = 0$ . $LHD = \lim_{h \to 0^-} \frac{0 - 1}{h} = \lim_{h \to 0^-} \frac{-1}{h}$ As $h$ approaches $0$ from the negative side, $\frac{-1}{h}$ approaches $+\infty$ . So, LHD $= +\infty$ . The limit does not exist as a finite number.
Right-Hand Derivative (RHD) at $x=1$ : RHD $= \lim_{h \to 0^+} \frac{f(1+h) - f(1)}{h}$ $= \lim_{h \to 0^+} \frac{[1+h] - 1}{h}$ As $h \to 0^+$ , $h$ is a small positive number. This means $1 < 1+h < 2$ . For example, if $h=0.1$ , $1+h=1.1$ . The greatest integer less than or equal to $1+h$ is $1$ . So, $[1+h] = 1$ . $RHD = \lim_{h \to 0^+} \frac{1 - 1}{h} = \lim_{h \to 0^+} \frac{0}{h} = 0$

Since LHD

(=+\infty)

and RHD

(=0)

are not equal, the function

f(x)=[x]

is not differentiable at

x=1

.

Case 2: Differentiability at $x=2$ First, we find the value of the function at

x=2

:

f(2) = [2] = 2

.

Left-Hand Derivative (LHD) at $x=2$ : LHD $= \lim_{h \to 0^-} \frac{f(2+h) - f(2)}{h}$ $= \lim_{h \to 0^-} \frac{[2+h] - 2}{h}$ As $h \to 0^-$ , $h$ is a small negative number. This means $1 < 2+h < 2$ . The greatest integer less than or equal to $2+h$ is $1$ . So, $[2+h] = 1$ . $LHD = \lim_{h \to 0^-} \frac{1 - 2}{h} = \lim_{h \to 0^-} \frac{-1}{h} = +\infty$ The limit does not exist as a finite number.
Right-Hand Derivative (RHD) at $x=2$ : RHD $= \lim_{h \to 0^+} \frac{f(2+h) - f(2)}{h}$ $= \lim_{h \to 0^+} \frac{[2+h] - 2}{h}$ As $h \to 0^+$ , $h$ is a small positive number. This means $2 < 2+h < 3$ . The greatest integer less than or equal to $2+h$ is $2$ . So, $[2+h] = 2$ . $RHD = \lim_{h \to 0^+} \frac{2 - 2}{h} = \lim_{h \to 0^+} \frac{0}{h} = 0$

Since LHD

(=+\infty)

and RHD

(=0)

are not equal, the function

f(x)=[x]

is not differentiable at

x=2

.

Conclusion: We have shown that the function

f(x)=[x]

is not differentiable at

x=1

and

x=2

.

Hence Proved.

Q1EXERCISE 5.3

$2 x+3 y=\sin x$

Solution

Given: The equation

2x + 3y = \sin x

.

To Find: The derivative

\frac{dy}{dx}

.

Solution: We are given an implicit function of

y

in terms of

x

. To find

\frac{dy}{dx}

, we differentiate both sides of the equation with respect to

x

.

Given equation:

2x + 3y = \sin x

Differentiating both sides with respect to

x

:

\frac{d}{dx}(2x + 3y) = \frac{d}{dx}(\sin x)

Using the sum rule for differentiation on the left side:

\frac{d}{dx}(2x) + \frac{d}{dx}(3y) = \frac{d}{dx}(\sin x)

Now, we differentiate each term:

The derivative of $2x$ with respect to $x$ is $2$ .
The derivative of $3y$ with respect to $x$ is $3\frac{dy}{dx}$ (using the chain rule).
The derivative of $\sin x$ with respect to $x$ is $\cos x$ .

Substituting these derivatives back into the equation:

2 + 3\frac{dy}{dx} = \cos x

Now, we need to solve for

\frac{dy}{dx}

. Subtract 2 from both sides:

3\frac{dy}{dx} = \cos x - 2

Divide by 3:

\frac{dy}{dx} = \frac{\cos x - 2}{3}

Final Answer: The derivative

\frac{dy}{dx}

for the equation

2x + 3y = \sin x

is

\frac{\cos x - 2}{3}

.

Q2EXERCISE 5.3

$2 x+3 y=\sin y$

Solution

Given: The equation

2x + 3y = \sin y

.

To Find: The derivative

\frac{dy}{dx}

.

Solution: We are given the equation

2x + 3y = \sin y

. To find

\frac{dy}{dx}

, we differentiate both sides of the equation with respect to

x

.

\frac{d}{dx}(2x + 3y) = \frac{d}{dx}(\sin y)

Applying the sum rule for differentiation on the left side:

\frac{d}{dx}(2x) + \frac{d}{dx}(3y) = \frac{d}{dx}(\sin y)

Now, we differentiate each term:

2 + 3\frac{dy}{dx} = \cos y \cdot \frac{dy}{dx}

Here, we used the chain rule for the terms involving

y

, since

y

is a function of

x

.

To solve for

\frac{dy}{dx}

, we rearrange the equation to group the

\frac{dy}{dx}

terms on one side:

2 = \cos y \frac{dy}{dx} - 3\frac{dy}{dx}

Factor out

\frac{dy}{dx}

from the right side:

2 = (\cos y - 3)\frac{dy}{dx}

Finally, we solve for

\frac{dy}{dx}

:

\frac{dy}{dx} = \frac{2}{\cos y - 3}

Final Answer: The derivative of the given function is

\frac{dy}{dx} = \frac{2}{\cos y - 3}

.

Q3EXERCISE 5.3

$a x+b y^{2}=\cos y$

Solution

Given: The equation

ax + by^2 = \cos y

.

To Find: The derivative

\frac{dy}{dx}

.

Solution: We are given the equation

ax + by^2 = \cos y

. To find

\frac{dy}{dx}

, we differentiate both sides of the equation with respect to

x

.

\frac{d}{dx}(ax + by^2) = \frac{d}{dx}(\cos y)

Applying the sum rule for differentiation on the left side:

\frac{d}{dx}(ax) + \frac{d}{dx}(by^2) = \frac{d}{dx}(\cos y)

Now, we differentiate each term using the power rule and the chain rule for terms involving

y

:

a \cdot 1 + b \cdot (2y) \frac{dy}{dx} = -\sin y \cdot \frac{dy}{dx}

a + 2by\frac{dy}{dx} = -\sin y \frac{dy}{dx}

To solve for

\frac{dy}{dx}

, we rearrange the equation to group all terms with

\frac{dy}{dx}

on one side:

a = -\sin y \frac{dy}{dx} - 2by\frac{dy}{dx}

Factor out

\frac{dy}{dx}

from the right side:

a = (-\sin y - 2by)\frac{dy}{dx}

a = -(\sin y + 2by)\frac{dy}{dx}

Finally, we solve for

\frac{dy}{dx}

:

\frac{dy}{dx} = \frac{a}{-(\sin y + 2by)} = -\frac{a}{\sin y + 2by}

Final Answer: The derivative of the given function is

\frac{dy}{dx} = -\frac{a}{2by + \sin y}

.

Q4EXERCISE 5.3

$x y+y^{2}=\tan x+y$

Solution

Given: The equation

xy + y^2 = \tan x + y

.

To Find: The derivative

\frac{dy}{dx}

.

Solution: We are given the equation

xy + y^2 = \tan x + y

. To find

\frac{dy}{dx}

, we differentiate both sides of the equation with respect to

x

.

\frac{d}{dx}(xy + y^2) = \frac{d}{dx}(\tan x + y)

Applying the sum rule on both sides:

\frac{d}{dx}(xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(\tan x) + \frac{d}{dx}(y)

Now, we differentiate each term. For the term

xy

, we use the product rule, and for

y^2

, we use the chain rule.

[\frac{d}{dx}(x) \cdot y + x \cdot \frac{d}{dx}(y)] + 2y \frac{dy}{dx} = \sec^2 x + \frac{dy}{dx}

[1 \cdot y + x \frac{dy}{dx}] + 2y \frac{dy}{dx} = \sec^2 x + \frac{dy}{dx}

y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = \sec^2 x + \frac{dy}{dx}

To solve for

\frac{dy}{dx}

, we rearrange the equation to group all terms with

\frac{dy}{dx}

on one side and the other terms on the other side:

x\frac{dy}{dx} + 2y\frac{dy}{dx} - \frac{dy}{dx} = \sec^2 x - y

Factor out

\frac{dy}{dx}

from the left side:

(x + 2y - 1)\frac{dy}{dx} = \sec^2 x - y

Finally, we solve for

\frac{dy}{dx}

:

\frac{dy}{dx} = \frac{\sec^2 x - y}{x + 2y - 1}

Final Answer: The derivative of the given function is

\frac{dy}{dx} = \frac{\sec^2 x - y}{x + 2y - 1}

.

Q5EXERCISE 5.3

$x^{2}+x y+y^{2}=100$

Solution

Given: The equation

x^2 + xy + y^2 = 100

.

To Find: The derivative

\frac{dy}{dx}

.

Solution: We are given the equation

x^2 + xy + y^2 = 100

. To find

\frac{dy}{dx}

, we differentiate both sides of the equation with respect to

x

.

\frac{d}{dx}(x^2 + xy + y^2) = \frac{d}{dx}(100)

Applying the sum rule on the left side:

\frac{d}{dx}(x^2) + \frac{d}{dx}(xy) + \frac{d}{dx}(y^2) = 0

Now, we differentiate each term. For the term

xy

, we use the product rule, and for

y^2

, we use the chain rule.

2x + [\frac{d}{dx}(x) \cdot y + x \cdot \frac{d}{dx}(y)] + 2y \frac{dy}{dx} = 0

2x + [1 \cdot y + x \frac{dy}{dx}] + 2y \frac{dy}{dx} = 0

2x + y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0

To solve for

\frac{dy}{dx}

, we rearrange the equation to group all terms with

\frac{dy}{dx}

on one side and the other terms on the other side:

x\frac{dy}{dx} + 2y\frac{dy}{dx} = -2x - y

Factor out

\frac{dy}{dx}

from the left side:

(x + 2y)\frac{dy}{dx} = -(2x + y)

Finally, we solve for

\frac{dy}{dx}

:

\frac{dy}{dx} = -\frac{2x + y}{x + 2y}

Final Answer: The derivative of the given function is

\frac{dy}{dx} = -\frac{2x + y}{x + 2y}

.

Q6EXERCISE 5.3

$x^{3}+x^{2} y+x y^{2}+y^{3}=81$

Solution

Given: The equation

x^3 + x^2y + xy^2 + y^3 = 81

.

To Find: The derivative

\frac{dy}{dx}

.

Solution: We are given the equation

x^3 + x^2y + xy^2 + y^3 = 81

. To find

\frac{dy}{dx}

, we differentiate both sides of the equation with respect to

x

.

\frac{d}{dx}(x^3 + x^2y + xy^2 + y^3) = \frac{d}{dx}(81)

Applying the sum rule on the left side:

\frac{d}{dx}(x^3) + \frac{d}{dx}(x^2y) + \frac{d}{dx}(xy^2) + \frac{d}{dx}(y^3) = 0

Now, we differentiate each term using the appropriate rules (power, product, and chain rules):

$\frac{d}{dx}(x^3) = 3x^2$
$\frac{d}{dx}(x^2y) = \frac{d}{dx}(x^2) \cdot y + x^2 \cdot \frac{d}{dx}(y) = 2xy + x^2\frac{dy}{dx}$
$\frac{d}{dx}(xy^2) = \frac{d}{dx}(x) \cdot y^2 + x \cdot \frac{d}{dx}(y^2) = 1 \cdot y^2 + x \cdot (2y\frac{dy}{dx}) = y^2 + 2xy\frac{dy}{dx}$
$\frac{d}{dx}(y^3) = 3y^2\frac{dy}{dx}$

Substituting these back into the equation:

3x^2 + (2xy + x^2\frac{dy}{dx}) + (y^2 + 2xy\frac{dy}{dx}) + 3y^2\frac{dy}{dx} = 0

Now, we group the terms with

\frac{dy}{dx}

and the terms without it:

(3x^2 + 2xy + y^2) + (x^2\frac{dy}{dx} + 2xy\frac{dy}{dx} + 3y^2\frac{dy}{dx}) = 0

(3x^2 + 2xy + y^2) + (x^2 + 2xy + 3y^2)\frac{dy}{dx} = 0

Rearrange to solve for

\frac{dy}{dx}

:

(x^2 + 2xy + 3y^2)\frac{dy}{dx} = -(3x^2 + 2xy + y^2)

\frac{dy}{dx} = -\frac{3x^2 + 2xy + y^2}{x^2 + 2xy + 3y^2}

Final Answer: The derivative of the given function is

\frac{dy}{dx} = -\frac{3x^2 + 2xy + y^2}{x^2 + 2xy + 3y^2}

.

Q7EXERCISE 5.3

$\sin ^{2} y+\cos x y=\kappa$

Solution

Given: The equation is

\sin^2 y + \cos(xy) = \kappa

, where

\kappa

is a constant.

To Find:

\frac{dy}{dx}

Solution: We differentiate the given equation with respect to

x

term by term.

\frac{d}{dx}(\sin^2 y) + \frac{d}{dx}(\cos(xy)) = \frac{d}{dx}(\kappa)

For the first term, using the chain rule:

\frac{d}{dx}(\sin^2 y) = 2 \sin y \cdot \frac{d}{dx}(\sin y) = 2 \sin y \cos y \cdot \frac{dy}{dx} = \sin(2y) \frac{dy}{dx}

For the second term, using the chain rule and the product rule for

xy

:

\frac{d}{dx}(\cos(xy)) = -\sin(xy) \cdot \frac{d}{dx}(xy)

\frac{d}{dx}(xy) = 1 \cdot y + x \cdot \frac{dy}{dx} = y + x \frac{dy}{dx}

So,

\frac{d}{dx}(\cos(xy)) = -\sin(xy) \left(y + x \frac{dy}{dx}\right) = -y \sin(xy) - x \sin(xy) \frac{dy}{dx}

For the right-hand side, the derivative of a constant

\kappa

is 0:

\frac{d}{dx}(\kappa) = 0

Substituting these back into the differentiated equation:

\sin(2y) \frac{dy}{dx} - y \sin(xy) - x \sin(xy) \frac{dy}{dx} = 0

Now, we group the terms with

\frac{dy}{dx}

:

\left( \sin(2y) - x \sin(xy) \right) \frac{dy}{dx} = y \sin(xy)

Finally, we solve for

\frac{dy}{dx}

:

\frac{dy}{dx} = \frac{y \sin(xy)}{\sin(2y) - x \sin(xy)}

Final Answer: The derivative

\frac{dy}{dx}

is

\frac{y \sin(xy)}{\sin(2y) - x \sin(xy)}

.

Q8EXERCISE 5.3

$\sin ^{2} x+\cos ^{2} y=1$

Solution

Given: The equation is

\sin^2 x + \cos^2 y = 1

.

To Find:

\frac{dy}{dx}

Solution: We differentiate the given equation with respect to

x

term by term.

\frac{d}{dx}(\sin^2 x) + \frac{d}{dx}(\cos^2 y) = \frac{d}{dx}(1)

For the first term, using the chain rule:

\frac{d}{dx}(\sin^2 x) = 2 \sin x \cdot \frac{d}{dx}(\sin x) = 2 \sin x \cos x = \sin(2x)

For the second term, using the chain rule:

\frac{d}{dx}(\cos^2 y) = 2 \cos y \cdot \frac{d}{dx}(\cos y) = 2 \cos y \cdot (-\sin y) \cdot \frac{dy}{dx} = -2 \sin y \cos y \frac{dy}{dx} = -\sin(2y) \frac{dy}{dx}

For the right-hand side, the derivative of the constant 1 is 0:

\frac{d}{dx}(1) = 0

Substituting these back into the differentiated equation:

\sin(2x) - \sin(2y) \frac{dy}{dx} = 0

Now, we solve for

\frac{dy}{dx}

:

\sin(2y) \frac{dy}{dx} = \sin(2x)

\frac{dy}{dx} = \frac{\sin(2x)}{\sin(2y)}

Final Answer: The derivative

\frac{dy}{dx}

is

\frac{\sin(2x)}{\sin(2y)}

.

Q9EXERCISE 5.3

$y=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$

Solution

Given: The function is

y = \sin^{-1}\left(\frac{2x}{1+x^2}\right)

.

To Find:

\frac{dy}{dx}

Solution: To simplify the expression, we use the substitution

x = \tan \theta

. This implies

\theta = \tan^{-1} x

.

Substituting

x = \tan \theta

into the expression inside the inverse sine function:

\frac{2x}{1+x^2} = \frac{2 \tan \theta}{1 + \tan^2 \theta}

We know the trigonometric identity

\sin(2\theta) = \frac{2 \tan \theta}{1 + \tan^2 \theta}

. So, the function becomes:

y = \sin^{-1}(\sin(2\theta))

Assuming that

2\theta

lies in the principal value branch of

\sin^{-1}

, which is

[-\frac{\pi}{2}, \frac{\pi}{2}]

, we can write

y = 2\theta

. This holds for

-1 \le x \le 1

.

Substitute back

\theta = \tan^{-1} x

:

y = 2 \tan^{-1} x

Now, we differentiate

y

with respect to

x

:

\frac{dy}{dx} = \frac{d}{dx}(2 \tan^{-1} x)

\frac{dy}{dx} = 2 \cdot \frac{d}{dx}(\tan^{-1} x)

We know that

\frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2}

.

\frac{dy}{dx} = 2 \cdot \frac{1}{1+x^2} = \frac{2}{1+x^2}

Final Answer: The derivative

\frac{dy}{dx}

is

\frac{2}{1+x^2}

.

Q10EXERCISE 5.3

$y=\tan ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right),-\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}$

Solution

Given: The function is

y = \tan^{-1}\left(\frac{3x-x^3}{1-3x^2}\right)

, for

-\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}

.

To Find:

\frac{dy}{dx}

Solution: To simplify the expression, we use the substitution

x = \tan \theta

. This implies

\theta = \tan^{-1} x

.

Substituting

x = \tan \theta

into the expression inside the inverse tangent function:

\frac{3x-x^3}{1-3x^2} = \frac{3\tan \theta - \tan^3 \theta}{1-3\tan^2 \theta}

We know the trigonometric identity

\tan(3\theta) = \frac{3\tan \theta - \tan^3 \theta}{1-3\tan^2 \theta}

. So, the function becomes:

y = \tan^{-1}(\tan(3\theta))

Now, we check the given domain for

x

:

-\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}

. Since

x = \tan \theta

:

\tan(-\frac{\pi}{6}) < \tan \theta < \tan(\frac{\pi}{6})

Since

\tan \theta

is an increasing function, this implies:

-\frac{\pi}{6} < \theta < \frac{\pi}{6}

Multiplying by 3, we get:

-\frac{3\pi}{6} < 3\theta < \frac{3\pi}{6} \implies -\frac{\pi}{2} < 3\theta < \frac{\pi}{2}

This range for

3\theta

is the principal value branch for

\tan^{-1}

. Therefore, we can write:

y = 3\theta

Substitute back

\theta = \tan^{-1} x

:

y = 3 \tan^{-1} x

Now, we differentiate

y

with respect to

x

:

\frac{dy}{dx} = \frac{d}{dx}(3 \tan^{-1} x)

\frac{dy}{dx} = 3 \cdot \frac{1}{1+x^2} = \frac{3}{1+x^2}

Final Answer: The derivative

\frac{dy}{dx}

is

\frac{3}{1+x^2}

.

Q11EXERCISE 5.3

$y=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right), 0<x<1$

Solution

Given: The function is

y = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)

, for

0 < x < 1

.

To Find:

\frac{dy}{dx}

Solution: To simplify the expression, we use the substitution

x = \tan \theta

. This implies

\theta = \tan^{-1} x

.

Substituting

x = \tan \theta

into the expression inside the inverse cosine function:

\frac{1-x^2}{1+x^2} = \frac{1-\tan^2 \theta}{1+\tan^2 \theta}

We know the trigonometric identity

\cos(2\theta) = \frac{1-\tan^2 \theta}{1+\tan^2 \theta}

. So, the function becomes:

y = \cos^{-1}(\cos(2\theta))

Now, we check the given domain for

x

:

0 < x < 1

. Since

x = \tan \theta

:

\tan(0) < \tan \theta < \tan(\frac{\pi}{4})

Since

\tan \theta

is an increasing function, this implies:

0 < \theta < \frac{\pi}{4}

Multiplying by 2, we get:

0 < 2\theta < \frac{\pi}{2}

This range for

2\theta

is within the principal value branch of

\cos^{-1}

, which is

[0, \pi]

. Therefore, we can write:

y = 2\theta

Substitute back

\theta = \tan^{-1} x

:

y = 2 \tan^{-1} x

Now, we differentiate

y

with respect to

x

:

\frac{dy}{dx} = \frac{d}{dx}(2 \tan^{-1} x)

\frac{dy}{dx} = 2 \cdot \frac{1}{1+x^2} = \frac{2}{1+x^2}

Final Answer: The derivative

\frac{dy}{dx}

is

\frac{2}{1+x^2}

.

Q12EXERCISE 5.3

$y=\sin ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right), 0<x<1$

Solution

Given: The function

y=\sin ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)

for

0<x<1

.

To Find: The derivative

\frac{dy}{dx}

.

Solution: Let the given function be

y=\sin ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)

. To simplify this expression, we use a trigonometric substitution. Let

x = \tan \theta

. This implies

\theta = \tan^{-1} x

.

Since

0 < x < 1

, we have

0 < \tan \theta < 1

. This means

0 < \theta < \frac{\pi}{4}

.

Substituting

x = \tan \theta

into the expression for

y

:

y = \sin^{-1}\left(\frac{1-\tan^2 \theta}{1+\tan^2 \theta}\right)

Using the trigonometric identity

\cos(2\theta) = \frac{1-\tan^2 \theta}{1+\tan^2 \theta}

, we get:

y = \sin^{-1}(\cos(2\theta))

We can write

\cos(2\theta)

as

\sin\left(\frac{\pi}{2} - 2\theta\right)

.

y = \sin^{-1}\left(\sin\left(\frac{\pi}{2} - 2\theta\right)\right)

Now, we check the range of the argument

\frac{\pi}{2} - 2\theta

. Since

0 < \theta < \frac{\pi}{4}

, we have

0 < 2\theta < \frac{\pi}{2}

. Multiplying by -1, we get

-\frac{\pi}{2} < -2\theta < 0

. Adding

\frac{\pi}{2}

to all parts of the inequality:

\frac{\pi}{2} - \frac{\pi}{2} < \frac{\pi}{2} - 2\theta < \frac{\pi}{2} + 0

0 < \frac{\pi}{2} - 2\theta < \frac{\pi}{2}

Since this range is within the principal value branch of

\sin^{-1}

, which is

\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

, we can write:

y = \frac{\pi}{2} - 2\theta

Substitute back

\theta = \tan^{-1} x

:

y = \frac{\pi}{2} - 2\tan^{-1} x

Now, we differentiate

y

with respect to

x

:

\frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{2} - 2\tan^{-1} x\right)

\frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{2}\right) - 2\frac{d}{dx}(\tan^{-1} x)

\frac{dy}{dx} = 0 - 2\left(\frac{1}{1+x^2}\right)

\frac{dy}{dx} = -\frac{2}{1+x^2}

Final Answer: The derivative of

y=\sin ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)

for

0<x<1

is

\frac{dy}{dx} = -\frac{2}{1+x^2}

.

Q13EXERCISE 5.3

$y=\cos ^{-1}\left(\frac{2 x}{1+x^{2}}\right),-1<x<1$

Solution

Given: The function

y=\cos ^{-1}\left(\frac{2 x}{1+x^{2}}\right)

for

-1<x<1

.

To Find: The derivative

\frac{dy}{dx}

.

Solution: Let the given function be

y=\cos ^{-1}\left(\frac{2 x}{1+x^{2}}\right)

. To simplify this expression, we use a trigonometric substitution. Let

x = \tan \theta

. This implies

\theta = \tan^{-1} x

.

Since

-1 < x < 1

, we have

-1 < \tan \theta < 1

. This means

-\frac{\pi}{4} < \theta < \frac{\pi}{4}

.

Substituting

x = \tan \theta

into the expression for

y

:

y = \cos^{-1}\left(\frac{2\tan \theta}{1+\tan^2 \theta}\right)

Using the trigonometric identity

\sin(2\theta) = \frac{2\tan \theta}{1+\tan^2 \theta}

, we get:

y = \cos^{-1}(\sin(2\theta))

We can write

\sin(2\theta)

as

\cos\left(\frac{\pi}{2} - 2\theta\right)

.

y = \cos^{-1}\left(\cos\left(\frac{\pi}{2} - 2\theta\right)\right)

Now, we check the range of the argument

\frac{\pi}{2} - 2\theta

. Since

-\frac{\pi}{4} < \theta < \frac{\pi}{4}

, we have

-\frac{\pi}{2} < 2\theta < \frac{\pi}{2}

. Multiplying by -1, we get

-\frac{\pi}{2} < -2\theta < \frac{\pi}{2}

. Adding

\frac{\pi}{2}

to all parts of the inequality:

\frac{\pi}{2} - \frac{\pi}{2} < \frac{\pi}{2} - 2\theta < \frac{\pi}{2} + \frac{\pi}{2}

0 < \frac{\pi}{2} - 2\theta < \pi

Since this range is within the principal value branch of

\cos^{-1}

, which is

[0, \pi]

, we can write:

y = \frac{\pi}{2} - 2\theta

Substitute back

\theta = \tan^{-1} x

:

y = \frac{\pi}{2} - 2\tan^{-1} x

Now, we differentiate

y

with respect to

x

:

\frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{2} - 2\tan^{-1} x\right)

\frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{2}\right) - 2\frac{d}{dx}(\tan^{-1} x)

\frac{dy}{dx} = 0 - 2\left(\frac{1}{1+x^2}\right)

\frac{dy}{dx} = -\frac{2}{1+x^2}

Final Answer: The derivative of

y=\cos ^{-1}\left(\frac{2 x}{1+x^{2}}\right)

for

-1<x<1

is

\frac{dy}{dx} = -\frac{2}{1+x^2}

.

Q14EXERCISE 5.3

$y=\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right),-\frac{1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}$

Solution

Given: The function

y=\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)

for

-\frac{1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}

.

To Find: The derivative

\frac{dy}{dx}

.

Solution: Let the given function be

y=\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)

. To simplify this expression, we use a trigonometric substitution. Let

x = \sin \theta

. This implies

\theta = \sin^{-1} x

.

Since

-\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}

, we have

-\frac{1}{\sqrt{2}} < \sin \theta < \frac{1}{\sqrt{2}}

. This means

-\frac{\pi}{4} < \theta < \frac{\pi}{4}

.

Substituting

x = \sin \theta

into the expression for

y

:

y = \sin^{-1}\left(2 \sin \theta \sqrt{1-\sin^2 \theta}\right)

For the given range of

\theta

,

\cos \theta

is positive, so

\sqrt{1-\sin^2 \theta} = \sqrt{\cos^2 \theta} = \cos \theta

.

y = \sin^{-1}(2 \sin \theta \cos \theta)

Using the trigonometric identity

\sin(2\theta) = 2 \sin \theta \cos \theta

, we get:

y = \sin^{-1}(\sin(2\theta))

Now, we check the range of the argument

2\theta

. Since

-\frac{\pi}{4} < \theta < \frac{\pi}{4}

, we have

-\frac{\pi}{2} < 2\theta < \frac{\pi}{2}

. Since this range is within the principal value branch of

\sin^{-1}

, which is

\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

, we can write:

y = 2\theta

Substitute back

\theta = \sin^{-1} x

:

y = 2\sin^{-1} x

Now, we differentiate

y

with respect to

x

:

\frac{dy}{dx} = \frac{d}{dx}(2\sin^{-1} x)

\frac{dy}{dx} = 2\frac{d}{dx}(\sin^{-1} x)

\frac{dy}{dx} = 2\left(\frac{1}{\sqrt{1-x^2}}\right)

\frac{dy}{dx} = \frac{2}{\sqrt{1-x^2}}

Final Answer: The derivative of

y=\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)

for

-\frac{1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}

is

\frac{dy}{dx} = \frac{2}{\sqrt{1-x^2}}

.

Q15EXERCISE 5.3

$y=\sec ^{-1}\left(\frac{1}{2 x^{2}-1}\right), 0<x<\frac{1}{\sqrt{2}}$

Solution

Given: The function

y=\sec ^{-1}\left(\frac{1}{2 x^{2}-1}\right)

for

0<x<\frac{1}{\sqrt{2}}

.

To Find: The derivative

\frac{dy}{dx}

.

Solution: Let the given function be

y=\sec ^{-1}\left(\frac{1}{2 x^{2}-1}\right)

. To simplify this expression, we use a trigonometric substitution. Let

x = \cos \theta

. This implies

\theta = \cos^{-1} x

.

Since

0 < x < \frac{1}{\sqrt{2}}

, we have

0 < \cos \theta < \frac{1}{\sqrt{2}}

. This means

\frac{\pi}{4} < \theta < \frac{\pi}{2}

.

Substituting

x = \cos \theta

into the expression for

y

:

y = \sec^{-1}\left(\frac{1}{2\cos^2 \theta - 1}\right)

Using the trigonometric identity

\cos(2\theta) = 2\cos^2 \theta - 1

, we get:

y = \sec^{-1}\left(\frac{1}{\cos(2\theta)}\right)

y = \sec^{-1}(\sec(2\theta))

Now, we check the range of the argument

2\theta

. Since

\frac{\pi}{4} < \theta < \frac{\pi}{2}

, we have

2\left(\frac{\pi}{4}\right) < 2\theta < 2\left(\frac{\pi}{2}\right)

, which simplifies to

\frac{\pi}{2} < 2\theta < \pi

. Since this range is within the principal value branch of

\sec^{-1}

, which is

[0, \pi] - \{\frac{\pi}{2}\}

, we can write:

y = 2\theta

Substitute back

\theta = \cos^{-1} x

:

y = 2\cos^{-1} x

Now, we differentiate

y

with respect to

x

:

\frac{dy}{dx} = \frac{d}{dx}(2\cos^{-1} x)

\frac{dy}{dx} = 2\frac{d}{dx}(\cos^{-1} x)

\frac{dy}{dx} = 2\left(\frac{-1}{\sqrt{1-x^2}}\right)

\frac{dy}{dx} = -\frac{2}{\sqrt{1-x^2}}

Final Answer: The derivative of

y=\sec ^{-1}\left(\frac{1}{2 x^{2}-1}\right)

for

0<x<\frac{1}{\sqrt{2}}

is

\frac{dy}{dx} = -\frac{2}{\sqrt{1-x^2}}

.

Q1EXERCISE 5.4

$\frac{e^{x}}{\sin x}$

Solution

Given: The function

y = \frac{e^{x}}{\sin x}

.

To Find: The derivative of the function with respect to

x

.

Formula: We will use the quotient rule for differentiation, which states that if

y = \frac{u(x)}{v(x)}

, then

\frac{dy}{dx} = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{[v(x)]^2}

Solution: Let the given function be

y = \frac{e^{x}}{\sin x}

. Here, let

u(x) = e^x

and

v(x) = \sin x

.

First, we find the derivatives of

u(x)

and

v(x)

:

u'(x) = \frac{d}{dx}(e^x) = e^x

v'(x) = \frac{d}{dx}(\sin x) = \cos x

Now, applying the quotient rule:

\frac{dy}{dx} = \frac{(\sin x) \cdot \frac{d}{dx}(e^x) - (e^x) \cdot \frac{d}{dx}(\sin x)}{(\sin x)^2}

\frac{dy}{dx} = \frac{(\sin x)(e^x) - (e^x)(\cos x)}{\sin^2 x}

Factoring out

e^x

from the numerator:

\frac{dy}{dx} = \frac{e^x(\sin x - \cos x)}{\sin^2 x}

This is the derivative of the given function, provided

\sin x \neq 0

.

Final Answer: The derivative of

\frac{e^{x}}{\sin x}

with respect to

x

is

\frac{e^x(\sin x - \cos x)}{\sin^2 x}

.

Q2EXERCISE 5.4

$e^{\sin ^{-1} x}$

Solution

Given: The function

y = e^{\sin^{-1} x}

.

To Find: The derivative of the function with respect to

x

, i.e.,

\frac{dy}{dx}

.

Solution: We use the chain rule for differentiation. Let

u = \sin^{-1} x

. Then the function becomes

y = e^u

.

The chain rule states that

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

.

First, we differentiate

y

with respect to

u

:

\frac{dy}{du} = \frac{d}{du}(e^u) = e^u = e^{\sin^{-1} x}

Next, we differentiate

u

with respect to

x

: We know that the derivative of

\sin^{-1} x

is

\frac{1}{\sqrt{1 - x^2}}

.

\frac{du}{dx} = \frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1 - x^2}}

Now, we multiply these derivatives to find

\frac{dy}{dx}

:

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = e^{\sin^{-1} x} \cdot \frac{1}{\sqrt{1 - x^2}}

\frac{dy}{dx} = \frac{e^{\sin^{-1} x}}{\sqrt{1 - x^2}}

Final Answer: The derivative of

e^{\sin^{-1} x}

with respect to

x

is

\frac{e^{\sin^{-1} x}}{\sqrt{1 - x^2}}

.

Q3EXERCISE 5.4

$e^{x^{3}}$

Solution

Given: The function

y = e^{x^3}

.

To Find: The derivative of the function with respect to

x

, i.e.,

\frac{dy}{dx}

.

Solution: We use the chain rule for differentiation. Let

u = x^3

. Then the function becomes

y = e^u

.

The chain rule states that

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

.

First, we differentiate

y

with respect to

u

:

\frac{dy}{du} = \frac{d}{du}(e^u) = e^u = e^{x^3}

Next, we differentiate

u

with respect to

x

using the power rule:

\frac{du}{dx} = \frac{d}{dx}(x^3) = 3x^2

Now, we multiply these derivatives to find

\frac{dy}{dx}

:

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = e^{x^3} \cdot (3x^2)

\frac{dy}{dx} = 3x^2 e^{x^3}

Final Answer: The derivative of

e^{x^3}

with respect to

x

is

3x^2 e^{x^3}

.

Q4EXERCISE 5.4

$\sin (\tan ^{-1} e^{-x})$

Solution

Given: The function

y = \sin(\tan^{-1} e^{-x})

.

To Find: The derivative of the function with respect to

x

, i.e.,

\frac{dy}{dx}

.

Solution: We apply the chain rule multiple times. Let

u = \tan^{-1} e^{-x}

and

v = e^{-x}

. Then

y = \sin(u)

and

u = \tan^{-1}(v)

.

The chain rule states that

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx}

.

Step 1: Differentiate

y

with respect to

u

.

\frac{dy}{du} = \frac{d}{du}(\sin u) = \cos u = \cos(\tan^{-1} e^{-x})

Step 2: Differentiate

u

with respect to

v

.

\frac{du}{dv} = \frac{d}{dv}(\tan^{-1} v) = \frac{1}{1+v^2} = \frac{1}{1+(e^{-x})^2} = \frac{1}{1+e^{-2x}}

Step 3: Differentiate

v

with respect to

x

.

\frac{dv}{dx} = \frac{d}{dx}(e^{-x}) = e^{-x} \cdot \frac{d}{dx}(-x) = e^{-x} \cdot (-1) = -e^{-x}

Step 4: Combine the results using the chain rule.

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx}

\frac{dy}{dx} = \cos(\tan^{-1} e^{-x}) \cdot \frac{1}{1+e^{-2x}} \cdot (-e^{-x})

\frac{dy}{dx} = -\frac{e^{-x} \cos(\tan^{-1} e^{-x})}{1+e^{-2x}}

Final Answer: The derivative of

\sin(\tan^{-1} e^{-x})

with respect to

x

is

-\frac{e^{-x} \cos(\tan^{-1} e^{-x})}{1+e^{-2x}}

.

Q5EXERCISE 5.4

$\log (\cos e^{x})$

Solution

Given: The function

y = \log(\cos e^x)

.

To Find: The derivative of the function with respect to

x

, i.e.,

\frac{dy}{dx}

.

Solution: We apply the chain rule. Let

u = \cos e^x

and

v = e^x

. Then

y = \log(u)

and

u = \cos(v)

.

The chain rule states that

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx}

.

Step 1: Differentiate

y

with respect to

u

.

\frac{dy}{du} = \frac{d}{du}(\log u) = \frac{1}{u} = \frac{1}{\cos(e^x)}

Step 2: Differentiate

u

with respect to

v

.

\frac{du}{dv} = \frac{d}{dv}(\cos v) = -\sin v = -\sin(e^x)

Step 3: Differentiate

v

with respect to

x

.

\frac{dv}{dx} = \frac{d}{dx}(e^x) = e^x

Step 4: Combine the results using the chain rule.

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx}

\frac{dy}{dx} = \frac{1}{\cos(e^x)} \cdot (-\sin(e^x)) \cdot e^x

\frac{dy}{dx} = -e^x \frac{\sin(e^x)}{\cos(e^x)}

Since

\frac{\sin \theta}{\cos \theta} = \tan \theta

, we have:

\frac{dy}{dx} = -e^x \tan(e^x)

Final Answer: The derivative of

\log(\cos e^x)

with respect to

x

is

-e^x \tan(e^x)

.

Q6EXERCISE 5.4

$e^{x}+e^{x^{2}}+\ldots+e^{x^{5}}$

Solution

Given: The function

y = e^x + e^{x^2} + \ldots + e^{x^5}

. This can be written explicitly as

y = e^x + e^{x^2} + e^{x^3} + e^{x^4} + e^{x^5}

.

To Find: The derivative of the function with respect to

x

, i.e.,

\frac{dy}{dx}

.

Solution: We use the sum rule for differentiation, which states that the derivative of a sum of functions is the sum of their derivatives.

\frac{dy}{dx} = \frac{d}{dx} (e^x + e^{x^2} + e^{x^3} + e^{x^4} + e^{x^5})

\frac{dy}{dx} = \frac{d}{dx}(e^x) + \frac{d}{dx}(e^{x^2}) + \frac{d}{dx}(e^{x^3}) + \frac{d}{dx}(e^{x^4}) + \frac{d}{dx}(e^{x^5})

Now we differentiate each term separately using the chain rule

\frac{d}{dx}(e^{f(x)}) = e^{f(x)} \cdot f'(x)

.

$\frac{d}{dx}(e^x) = e^x$
$\frac{d}{dx}(e^{x^2}) = e^{x^2} \cdot \frac{d}{dx}(x^2) = e^{x^2} \cdot (2x) = 2x e^{x^2}$
$\frac{d}{dx}(e^{x^3}) = e^{x^3} \cdot \frac{d}{dx}(x^3) = e^{x^3} \cdot (3x^2) = 3x^2 e^{x^3}$
$\frac{d}{dx}(e^{x^4}) = e^{x^4} \cdot \frac{d}{dx}(x^4) = e^{x^4} \cdot (4x^3) = 4x^3 e^{x^4}$
$\frac{d}{dx}(e^{x^5}) = e^{x^5} \cdot \frac{d}{dx}(x^5) = e^{x^5} \cdot (5x^4) = 5x^4 e^{x^5}$

Adding all the derivatives together:

\frac{dy}{dx} = e^x + 2x e^{x^2} + 3x^2 e^{x^3} + 4x^3 e^{x^4} + 5x^4 e^{x^5}

Final Answer: The derivative of

e^x + e^{x^2} + \ldots + e^{x^5}

with respect to

x

is

e^x + 2x e^{x^2} + 3x^2 e^{x^3} + 4x^3 e^{x^4} + 5x^4 e^{x^5}

.

Q7EXERCISE 5.4

$\sqrt{e^{\sqrt{x}}}, x>0$

Solution

Given: The function

y = \sqrt{e^{\sqrt{x}}}

for

x > 0

.

To Find: The derivative of the function with respect to

x

.

Solution: Let the given function be

y = \sqrt{e^{\sqrt{x}}}

. This can be written as

y = (e^{\sqrt{x}})^{1/2} = e^{\frac{1}{2}\sqrt{x}}

.

We will use the chain rule to differentiate

y

with respect to

x

. The chain rule states that if

y = f(g(h(x)))

, then

\frac{dy}{dx} = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x)

.

Let's apply this step-by-step:

y = e^u

, where

u = \frac{1}{2}v

, and

v = \sqrt{x} = x^{1/2}

.

First, differentiate

y

with respect to

u

:

\frac{dy}{du} = \frac{d}{du}(e^u) = e^u = e^{\frac{1}{2}\sqrt{x}} = \sqrt{e^{\sqrt{x}}}

.

Next, differentiate

u

with respect to

v

:

\frac{du}{dv} = \frac{d}{dv}\left(\frac{1}{2}v\right) = \frac{1}{2}

.

Finally, differentiate

v

with respect to

x

:

\frac{dv}{dx} = \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}

.

Now, by the chain rule,

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx}

.

\frac{dy}{dx} = e^{\frac{1}{2}\sqrt{x}} \cdot \frac{1}{2} \cdot \frac{1}{2\sqrt{x}}

\frac{dy}{dx} = \frac{e^{\frac{1}{2}\sqrt{x}}}{4\sqrt{x}}

Substituting back

e^{\frac{1}{2}\sqrt{x}} = \sqrt{e^{\sqrt{x}}}

:

\frac{dy}{dx} = \frac{\sqrt{e^{\sqrt{x}}}}{4\sqrt{x}}

Final Answer: The derivative of

\sqrt{e^{\sqrt{x}}}

with respect to

x

is

\frac{\sqrt{e^{\sqrt{x}}}}{4\sqrt{x}}

.

Q8EXERCISE 5.4

$\log (\log x), x>1$

Solution

Given: The function

y = \log(\log x)

for

x > 1

.

To Find: The derivative of the function with respect to

x

.

Solution: Let the given function be

y = \log(\log x)

. We will use the chain rule to differentiate

y

with respect to

x

. The chain rule states that if

y = f(g(x))

, then

\frac{dy}{dx} = f'(g(x)) \cdot g'(x)

.

Let

u = \log x

. Then

y = \log u

.

First, differentiate

y

with respect to

u

:

\frac{dy}{du} = \frac{d}{du}(\log u) = \frac{1}{u}

Substituting back

u = \log x

, we get:

\frac{dy}{du} = \frac{1}{\log x}

Next, differentiate

u

with respect to

x

:

\frac{du}{dx} = \frac{d}{dx}(\log x) = \frac{1}{x}

Now, by the chain rule,

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

.

\frac{dy}{dx} = \frac{1}{\log x} \cdot \frac{1}{x}

\frac{dy}{dx} = \frac{1}{x \log x}

The condition

x > 1

ensures that

\log x > 0

, so the function

\log(\log x)

is well-defined.

Final Answer: The derivative of

\log(\log x)

with respect to

x

is

\frac{1}{x \log x}

.

Q9EXERCISE 5.4

$\frac{\cos x}{\log x}, x>0$

Solution

Given: The function

y = \frac{\cos x}{\log x}

for

x > 0

.

To Find: The derivative of the function with respect to

x

.

Solution: Let the given function be

y = \frac{\cos x}{\log x}

. This is a quotient of two functions, so we will use the quotient rule for differentiation.

Formula: The quotient rule states that if

y = \frac{u(x)}{v(x)}

, then

\frac{dy}{dx} = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{[v(x)]^2}

.

Here, let

u(x) = \cos x

and

v(x) = \log x

.

First, find the derivatives of

u(x)

and

v(x)

:

u'(x) = \frac{d}{dx}(\cos x) = -\sin x

v'(x) = \frac{d}{dx}(\log x) = \frac{1}{x}

Now, apply the quotient rule:

\frac{dy}{dx} = \frac{(\log x)(-\sin x) - (\cos x)\left(\frac{1}{x}\right)}{(\log x)^2}

Simplify the numerator:

\frac{dy}{dx} = \frac{-\sin x \log x - \frac{\cos x}{x}}{(\log x)^2}

To remove the fraction in the numerator, we can multiply the numerator and denominator by

x

:

\frac{dy}{dx} = \frac{x(-\sin x \log x - \frac{\cos x}{x})}{x(\log x)^2}

\frac{dy}{dx} = \frac{-x \sin x \log x - \cos x}{x(\log x)^2}

We can factor out the negative sign from the numerator:

\frac{dy}{dx} = -\frac{x \sin x \log x + \cos x}{x(\log x)^2}

Final Answer: The derivative of

\frac{\cos x}{\log x}

with respect to

x

is

-\frac{x \sin x \log x + \cos x}{x(\log x)^2}

.

Q10EXERCISE 5.4

$\cos (\log x+e^{x}), x>0$

Solution

Given: The function

y = \cos(\log x + e^x)

for

x > 0

.

To Find: The derivative of the function with respect to

x

.

Solution: Let the given function be

y = \cos(\log x + e^x)

. We will use the chain rule to differentiate

y

with respect to

x

. The chain rule states that if

y = f(g(x))

, then

\frac{dy}{dx} = f'(g(x)) \cdot g'(x)

.

Let

u = \log x + e^x

. Then

y = \cos u

.

First, differentiate

y

with respect to

u

:

\frac{dy}{du} = \frac{d}{du}(\cos u) = -\sin u

Substituting back

u = \log x + e^x

, we get:

\frac{dy}{du} = -\sin(\log x + e^x)

Next, differentiate

u

with respect to

x

using the sum rule:

\frac{du}{dx} = \frac{d}{dx}(\log x + e^x) = \frac{d}{dx}(\log x) + \frac{d}{dx}(e^x)

\frac{du}{dx} = \frac{1}{x} + e^x

Now, by the chain rule,

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

.

\frac{dy}{dx} = [-\sin(\log x + e^x)] \cdot \left(\frac{1}{x} + e^x\right)

\frac{dy}{dx} = -\left(\frac{1}{x} + e^x\right) \sin(\log x + e^x)

Final Answer: The derivative of

\cos(\log x + e^x)

with respect to

x

is

-\left(\frac{1}{x} + e^x\right) \sin(\log x + e^x)

.

Q1EXERCISE 5.5

$\cos x \cdot \cos 2 x \cdot \cos 3 x$

Solution

Given: The function

y = \cos x \cdot \cos 2x \cdot \cos 3x

.

To Find: The derivative of the function with respect to

x

.

Solution: The function is a product of three functions. Differentiating it directly using the product rule is possible but can be lengthy. A more efficient method is logarithmic differentiation.

Let

y = \cos x \cdot \cos 2x \cdot \cos 3x

.

Step 1: Take the natural logarithm of both sides.

\log y = \log (\cos x \cdot \cos 2x \cdot \cos 3x)

Using the logarithm property

\log(abc) = \log a + \log b + \log c

:

\log y = \log(\cos x) + \log(\cos 2x) + \log(\cos 3x)

Step 2: Differentiate both sides with respect to

x

.

\frac{d}{dx}(\log y) = \frac{d}{dx}[\log(\cos x) + \log(\cos 2x) + \log(\cos 3x)]

Using the chain rule for each term:

\frac{1}{y} \frac{dy}{dx} = \frac{1}{\cos x} \cdot \frac{d}{dx}(\cos x) + \frac{1}{\cos 2x} \cdot \frac{d}{dx}(\cos 2x) + \frac{1}{\cos 3x} \cdot \frac{d}{dx}(\cos 3x)

\frac{1}{y} \frac{dy}{dx} = \frac{1}{\cos x} (-\sin x) + \frac{1}{\cos 2x} (-2\sin 2x) + \frac{1}{\cos 3x} (-3\sin 3x)

Recall that

\frac{\sin \theta}{\cos \theta} = \tan \theta

:

\frac{1}{y} \frac{dy}{dx} = -\tan x - 2\tan 2x - 3\tan 3x

\frac{1}{y} \frac{dy}{dx} = -(\tan x + 2\tan 2x + 3\tan 3x)

Step 3: Solve for

\frac{dy}{dx}

.

\frac{dy}{dx} = y \cdot [-(\tan x + 2\tan 2x + 3\tan 3x)]

Step 4: Substitute the original expression for

y

.

\frac{dy}{dx} = -(\cos x \cdot \cos 2x \cdot \cos 3x) (\tan x + 2\tan 2x + 3\tan 3x)

Final Answer: The derivative of

\cos x \cdot \cos 2x \cdot \cos 3x

is

-(\cos x \cdot \cos 2x \cdot \cos 3x)(\tan x + 2\tan 2x + 3\tan 3x)

.

Q2EXERCISE 5.5

$\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}$

Solution

Given: The function

y = \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}

.

To Find: The derivative of the function with respect to

x

.

Method: Since the function involves a product and quotient of several terms under a square root, logarithmic differentiation is the most suitable method.

Solution: Let

y = \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}} = \left(\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}\right)^{\frac{1}{2}}

.

Taking the natural logarithm on both sides:

\log y = \log \left[ \left(\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}\right)^{\frac{1}{2}} \right]

Using the logarithm property

\log a^b = b \log a

:

\log y = \frac{1}{2} \log \left[ \frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)} \right]

Using the properties

\log(\frac{a}{b}) = \log a - \log b

and

\log(ab) = \log a + \log b

:

\log y = \frac{1}{2} [\log((x-1)(x-2)) - \log((x-3)(x-4)(x-5))]

\log y = \frac{1}{2} [(\log(x-1) + \log(x-2)) - (\log(x-3) + \log(x-4) + \log(x-5))]

\log y = \frac{1}{2} [\log(x-1) + \log(x-2) - \log(x-3) - \log(x-4) - \log(x-5)]

Now, differentiate both sides with respect to

x

:

\frac{d}{dx}(\log y) = \frac{d}{dx} \left( \frac{1}{2} [\log(x-1) + \log(x-2) - \log(x-3) - \log(x-4) - \log(x-5)] \right)

\frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \left[ \frac{1}{x-1} + \frac{1}{x-2} - \frac{1}{x-3} - \frac{1}{x-4} - \frac{1}{x-5} \right]

To find

\frac{dy}{dx}

, multiply both sides by

y

:

\frac{dy}{dx} = y \cdot \frac{1}{2} \left[ \frac{1}{x-1} + \frac{1}{x-2} - \frac{1}{x-3} - \frac{1}{x-4} - \frac{1}{x-5} \right]

Substitute the original expression for

y

back into the equation:

\frac{dy}{dx} = \frac{1}{2} \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}} \left[ \frac{1}{x-1} + \frac{1}{x-2} - \frac{1}{x-3} - \frac{1}{x-4} - \frac{1}{x-5} \right]

Final Answer: The derivative of the given function is:

\frac{dy}{dx} = \frac{1}{2} \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}} \left[ \frac{1}{x-1} + \frac{1}{x-2} - \frac{1}{x-3} - \frac{1}{x-4} - \frac{1}{x-5} \right]

Q3EXERCISE 5.5

$(\log x)^{\cos x}$

Solution

Given: The function

y = (\log x)^{\cos x}

.

To Find: The derivative of the function with respect to

x

.

Method: The function is of the form

f(x)^{g(x)}

, so we use logarithmic differentiation.

Solution: Let

y = (\log x)^{\cos x}

.

Taking the natural logarithm on both sides:

\log y = \log((\log x)^{\cos x})

Using the logarithm property

\log a^b = b \log a

:

\log y = \cos x \cdot \log(\log x)

Now, differentiate both sides with respect to

x

. We use the product rule on the right-hand side, which states

(uv)' = u'v + uv'

. Let

u = \cos x

and

v = \log(\log x)

. Then

u' = -\sin x

. To find

v'

, we use the chain rule:

v' = \frac{d}{dx}(\log(\log x)) = \frac{1}{\log x} \cdot \frac{d}{dx}(\log x) = \frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x \log x}

.

Applying the product rule:

\frac{d}{dx}(\log y) = \frac{d}{dx}(\cos x \cdot \log(\log x))

\frac{1}{y} \frac{dy}{dx} = (\frac{d}{dx}\cos x) \cdot \log(\log x) + \cos x \cdot (\frac{d}{dx}\log(\log x))

\frac{1}{y} \frac{dy}{dx} = (-\sin x) \cdot \log(\log x) + \cos x \cdot \left( \frac{1}{x \log x} \right)

\frac{1}{y} \frac{dy}{dx} = \frac{\cos x}{x \log x} - \sin x \log(\log x)

To find

\frac{dy}{dx}

, multiply both sides by

y

:

\frac{dy}{dx} = y \left( \frac{\cos x}{x \log x} - \sin x \log(\log x) \right)

Substitute the original expression for

y

back into the equation:

\frac{dy}{dx} = (\log x)^{\cos x} \left( \frac{\cos x}{x \log x} - \sin x \log(\log x) \right)

Final Answer: The derivative of the given function is:

\frac{dy}{dx} = (\log x)^{\cos x} \left( \frac{\cos x}{x \log x} - \sin x \log(\log x) \right)

Q4EXERCISE 5.5

Differentiate the functions given in Exercises 1 to 11 w.r.t. $x$ . 4. $x^{x}-2^{\sin x}$

Solution

Given: The function

y = x^{x}-2^{\sin x}

.

To Find: The derivative of the function with respect to

x

.

Method: The function is a difference of two functions,

u(x) - v(x)

. We cannot take the logarithm of the entire expression directly. We must differentiate each part separately and then subtract the results. Let

y = u - v

, where

u = x^x

and

v = 2^{\sin x}

. Then

\frac{dy}{dx} = \frac{du}{dx} - \frac{dv}{dx}

.

Solution:

Part 1: Differentiate $u = x^x$ This is of the form

f(x)^{g(x)}

, so we use logarithmic differentiation.

u = x^x

Taking the natural logarithm:

\log u = \log(x^x) = x \log x

Differentiating with respect to

x

using the product rule:

\frac{1}{u} \frac{du}{dx} = \frac{d}{dx}(x) \cdot \log x + x \cdot \frac{d}{dx}(\log x)

\frac{1}{u} \frac{du}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x}

\frac{1}{u} \frac{du}{dx} = \log x + 1

\frac{du}{dx} = u(1 + \log x)

Substituting

u = x^x

:

\frac{du}{dx} = x^x(1 + \log x)

Part 2: Differentiate $v = 2^{\sin x}$ This is of the form

a^{f(x)}

. We can use the formula

\frac{d}{dx}(a^z) = a^z \log a \frac{dz}{dx}

or logarithmic differentiation. Using the formula with

a=2

and

z=\sin x

:

\frac{dv}{dx} = 2^{\sin x} \cdot \log 2 \cdot \frac{d}{dx}(\sin x)

\frac{dv}{dx} = 2^{\sin x} (\log 2) (\cos x)

Alternatively, using logarithmic differentiation for

v = 2^{\sin x}

:

\log v = \log(2^{\sin x}) = (\sin x) \log 2

Differentiating with respect to

x

:

\frac{1}{v} \frac{dv}{dx} = \frac{d}{dx}((\sin x) \log 2) = (\cos x) \log 2

\frac{dv}{dx} = v(\cos x \log 2)

Substituting

v = 2^{\sin x}

:

\frac{dv}{dx} = 2^{\sin x} \cos x \log 2

Combine the results: We have

\frac{dy}{dx} = \frac{du}{dx} - \frac{dv}{dx}

.

\frac{dy}{dx} = x^x(1 + \log x) - 2^{\sin x} \cos x \log 2

Final Answer: The derivative of the given function is:

\frac{dy}{dx} = x^x(1 + \log x) - 2^{\sin x} \cos x \log 2

Q5EXERCISE 5.5

Differentiate the functions given in Exercises 1 to 11 w.r.t. $x$ . 5. $(x+3)^{2}\cdot(x+4)^{3}\cdot(x+5)^{4}$

Solution

Given: The function

y = (x+3)^{2} \cdot (x+4)^{3} \cdot (x+5)^{4}

.

To Find: The derivative of the function with respect to

x

.

Method: Since the function is a product of several functions raised to powers, logarithmic differentiation is an efficient method.

Solution: Let

y = (x+3)^{2}(x+4)^{3}(x+5)^{4}

.

Taking the natural logarithm on both sides:

\log y = \log((x+3)^{2}(x+4)^{3}(x+5)^{4})

Using the logarithm property

\log(abc) = \log a + \log b + \log c

:

\log y = \log((x+3)^2) + \log((x+4)^3) + \log((x+5)^4)

Using the property

\log a^b = b \log a

:

\log y = 2\log(x+3) + 3\log(x+4) + 4\log(x+5)

Now, differentiate both sides with respect to

x

:

\frac{d}{dx}(\log y) = \frac{d}{dx} [2\log(x+3) + 3\log(x+4) + 4\log(x+5)]

\frac{1}{y} \frac{dy}{dx} = 2 \cdot \frac{1}{x+3} + 3 \cdot \frac{1}{x+4} + 4 \cdot \frac{1}{x+5}

\frac{1}{y} \frac{dy}{dx} = \frac{2}{x+3} + \frac{3}{x+4} + \frac{4}{x+5}

To find

\frac{dy}{dx}

, multiply both sides by

y

:

\frac{dy}{dx} = y \left( \frac{2}{x+3} + \frac{3}{x+4} + \frac{4}{x+5} \right)

Substitute the original expression for

y

back into the equation:

\frac{dy}{dx} = (x+3)^{2}(x+4)^{3}(x+5)^{4} \left( \frac{2}{x+3} + \frac{3}{x+4} + \frac{4}{x+5} \right)

To simplify the expression, we can find a common denominator for the terms in the parenthesis:

\frac{2}{x+3} + \frac{3}{x+4} + \frac{4}{x+5} = \frac{2(x+4)(x+5) + 3(x+3)(x+5) + 4(x+3)(x+4)}{(x+3)(x+4)(x+5)}

= \frac{2(x^2+9x+20) + 3(x^2+8x+15) + 4(x^2+7x+12)}{(x+3)(x+4)(x+5)}

= \frac{(2x^2+18x+40) + (3x^2+24x+45) + (4x^2+28x+48)}{(x+3)(x+4)(x+5)}

= \frac{(2+3+4)x^2 + (18+24+28)x + (40+45+48)}{(x+3)(x+4)(x+5)}

= \frac{9x^2 + 70x + 133}{(x+3)(x+4)(x+5)}

Now, substitute this back into the expression for

\frac{dy}{dx}

:

\frac{dy}{dx} = (x+3)^{2}(x+4)^{3}(x+5)^{4} \cdot \frac{9x^2 + 70x + 133}{(x+3)(x+4)(x+5)}

Cancelling common factors:

\frac{dy}{dx} = (x+3)^{2-1}(x+4)^{3-1}(x+5)^{4-1} (9x^2 + 70x + 133)

\frac{dy}{dx} = (x+3)(x+4)^{2}(x+5)^{3} (9x^2 + 70x + 133)

Final Answer: The derivative of the given function is:

\frac{dy}{dx} = (x+3)(x+4)^{2}(x+5)^{3} (9x^2 + 70x + 133)

Q6EXERCISE 5.5

Differentiate the functions given in Exercises 1 to 11 w.r.t. $x$ . 6. $\left(x+\frac{1}{x}\right)^{x}+x^{\left(1+\frac{1}{x}\right)}$

Solution

Given: The function

y = \left(x+\frac{1}{x}\right)^{x}+x^{\left(1+\frac{1}{x}\right)}

.

To Find: The derivative of the function with respect to

x

.

Method: The function is a sum of two functions,

u(x) + v(x)

. We cannot take the logarithm of the entire expression directly. We must differentiate each part separately and then add the results. Let

y = u + v

, where

u = \left(x+\frac{1}{x}\right)^{x}

and

v = x^{\left(1+\frac{1}{x}\right)}

. Then

\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}

.

Solution:

Part 1: Differentiate $u = \left(x+\frac{1}{x}\right)^{x}$ This is of the form

f(x)^{g(x)}

, so we use logarithmic differentiation.

u = \left(x+\frac{1}{x}\right)^{x}

Taking the natural logarithm:

\log u = \log\left(x+\frac{1}{x}\right)^{x} = x \log\left(x+\frac{1}{x}\right)

Differentiating with respect to

x

using the product rule:

\frac{1}{u} \frac{du}{dx} = \frac{d}{dx}(x) \cdot \log\left(x+\frac{1}{x}\right) + x \cdot \frac{d}{dx}\left(\log(x+\frac{1}{x})\right)

\frac{1}{u} \frac{du}{dx} = 1 \cdot \log\left(x+\frac{1}{x}\right) + x \cdot \frac{1}{x+\frac{1}{x}} \cdot \frac{d}{dx}\left(x+\frac{1}{x}\right)

\frac{1}{u} \frac{du}{dx} = \log\left(x+\frac{1}{x}\right) + x \cdot \frac{1}{\frac{x^2+1}{x}} \cdot \left(1 - \frac{1}{x^2}\right)

\frac{1}{u} \frac{du}{dx} = \log\left(x+\frac{1}{x}\right) + x \cdot \frac{x}{x^2+1} \cdot \left(\frac{x^2-1}{x^2}\right)

\frac{1}{u} \frac{du}{dx} = \log\left(x+\frac{1}{x}\right) + \frac{x^2}{x^2+1} \cdot \frac{x^2-1}{x^2}

\frac{1}{u} \frac{du}{dx} = \log\left(x+\frac{1}{x}\right) + \frac{x^2-1}{x^2+1}

\frac{du}{dx} = u \left[ \log\left(x+\frac{1}{x}\right) + \frac{x^2-1}{x^2+1} \right]

\frac{du}{dx} = \left(x+\frac{1}{x}\right)^{x} \left[ \log\left(x+\frac{1}{x}\right) + \frac{x^2-1}{x^2+1} \right]

Part 2: Differentiate $v = x^{\left(1+\frac{1}{x}\right)}$ This is also of the form

f(x)^{g(x)}

, so we use logarithmic differentiation.

v = x^{\left(1+\frac{1}{x}\right)}

Taking the natural logarithm:

\log v = \log\left(x^{\left(1+\frac{1}{x}\right)}\right) = \left(1+\frac{1}{x}\right) \log x

Differentiating with respect to

x

using the product rule:

\frac{1}{v} \frac{dv}{dx} = \frac{d}{dx}\left(1+\frac{1}{x}\right) \cdot \log x + \left(1+\frac{1}{x}\right) \cdot \frac{d}{dx}(\log x)

\frac{1}{v} \frac{dv}{dx} = \left(-\frac{1}{x^2}\right) \log x + \left(1+\frac{1}{x}\right) \cdot \frac{1}{x}

\frac{1}{v} \frac{dv}{dx} = -\frac{\log x}{x^2} + \frac{1}{x} + \frac{1}{x^2}

\frac{1}{v} \frac{dv}{dx} = \frac{x+1-\log x}{x^2}

\frac{dv}{dx} = v \left[ \frac{x+1-\log x}{x^2} \right]

\frac{dv}{dx} = x^{\left(1+\frac{1}{x}\right)} \frac{x+1-\log x}{x^2}

Combine the results: We have

\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}

.

\frac{dy}{dx} = \left(x+\frac{1}{x}\right)^{x} \left[ \log\left(x+\frac{1}{x}\right) + \frac{x^2-1}{x^2+1} \right] + x^{\left(1+\frac{1}{x}\right)} \frac{x+1-\log x}{x^2}

Final Answer: The derivative of the given function is:

\frac{dy}{dx} = \left(x+\frac{1}{x}\right)^{x} \left( \frac{x^2-1}{x^2+1} + \log\left(x+\frac{1}{x}\right) \right) + x^{1+\frac{1}{x}} \left( \frac{x+1-\log x}{x^2} \right)

Q7EXERCISE 5.5

Differentiate the functions given in Exercises 1 to 11 w.r.t. $x$ . 7. $(\log x)^{x}+x^{\log x}$

Solution

Given: The function

y = (\log x)^{x}+x^{\log x}

.

To Find: The derivative of the function with respect to

x

.

Solution: Let the given function be

y = u + v

, where

u = (\log x)^{x}

and

v = x^{\log x}

. Then,

\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}

.

First, consider

u = (\log x)^{x}

. This is of the form

f(x)^{g(x)}

, so we use logarithmic differentiation. Taking the natural logarithm on both sides:

\log u = \log ((\log x)^{x})

\log u = x \log(\log x)

Differentiating both sides with respect to

x

using the product rule:

\frac{1}{u} \frac{du}{dx} = \frac{d}{dx}(x) \cdot \log(\log x) + x \cdot \frac{d}{dx}(\log(\log x))

\frac{1}{u} \frac{du}{dx} = 1 \cdot \log(\log x) + x \cdot \frac{1}{\log x} \cdot \frac{d}{dx}(\log x)

\frac{1}{u} \frac{du}{dx} = \log(\log x) + x \cdot \frac{1}{\log x} \cdot \frac{1}{x}

\frac{1}{u} \frac{du}{dx} = \log(\log x) + \frac{1}{\log x}

\frac{du}{dx} = u \left( \log(\log x) + \frac{1}{\log x} \right)

Substituting back

u = (\log x)^{x}

:

\frac{du}{dx} = (\log x)^{x} \left( \log(\log x) + \frac{1}{\log x} \right) \quad ...(1)

Next, consider

v = x^{\log x}

. Taking the natural logarithm on both sides:

\log v = \log(x^{\log x})

\log v = (\log x)(\log x) = (\log x)^2

Differentiating both sides with respect to

x

using the chain rule:

\frac{1}{v} \frac{dv}{dx} = 2(\log x) \cdot \frac{d}{dx}(\log x)

\frac{1}{v} \frac{dv}{dx} = 2(\log x) \cdot \frac{1}{x}

\frac{dv}{dx} = v \left( \frac{2\log x}{x} \right)

Substituting back

v = x^{\log x}

:

\frac{dv}{dx} = x^{\log x} \left( \frac{2\log x}{x} \right) \quad ...(2)

Now, adding equations (1) and (2) to get

\frac{dy}{dx}

:

\frac{dy}{dx} = (\log x)^{x} \left( \log(\log x) + \frac{1}{\log x} \right) + x^{\log x} \cdot \frac{2\log x}{x}

Final Answer: The derivative of the given function is

(\log x)^{x} \left( \log(\log x) + \frac{1}{\log x} \right) + x^{\log x-1} \cdot 2\log x

.

Q8EXERCISE 5.5

Differentiate the functions given in Exercises 1 to 11 w.r.t. $x$ . 8. $(\sin x)^{x}+\sin ^{-1} \sqrt{x}$

Solution

Given: The function

y = (\sin x)^{x}+\sin ^{-1} \sqrt{x}

.

To Find: The derivative of the function with respect to

x

.

Solution: Let the given function be

y = u + v

, where

u = (\sin x)^{x}

and

v = \sin^{-1} \sqrt{x}

. Then,

\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}

.

First, consider

u = (\sin x)^{x}

. This is of the form

f(x)^{g(x)}

, so we use logarithmic differentiation. Taking the natural logarithm on both sides:

\log u = \log((\sin x)^{x})

\log u = x \log(\sin x)

Differentiating both sides with respect to

x

using the product rule:

\frac{1}{u} \frac{du}{dx} = \frac{d}{dx}(x) \cdot \log(\sin x) + x \cdot \frac{d}{dx}(\log(\sin x))

\frac{1}{u} \frac{du}{dx} = 1 \cdot \log(\sin x) + x \cdot \frac{1}{\sin x} \cdot \cos x

\frac{1}{u} \frac{du}{dx} = \log(\sin x) + x \cot x

\frac{du}{dx} = u (\log(\sin x) + x \cot x)

Substituting back

u = (\sin x)^{x}

:

\frac{du}{dx} = (\sin x)^{x} (\log(\sin x) + x \cot x) \quad ...(1)

Next, consider

v = \sin^{-1} \sqrt{x}

. We differentiate this using the chain rule. The derivative of

\sin^{-1} z

is

\frac{1}{\sqrt{1-z^2}}

.

\frac{dv}{dx} = \frac{d}{dx}(\sin^{-1} \sqrt{x})

\frac{dv}{dx} = \frac{1}{\sqrt{1 - (\sqrt{x})^2}} \cdot \frac{d}{dx}(\sqrt{x})

\frac{dv}{dx} = \frac{1}{\sqrt{1-x}} \cdot \frac{1}{2\sqrt{x}}

\frac{dv}{dx} = \frac{1}{2\sqrt{x(1-x)}} = \frac{1}{2\sqrt{x-x^2}} \quad ...(2)

Now, adding equations (1) and (2) to get

\frac{dy}{dx}

:

\frac{dy}{dx} = (\sin x)^{x} (\log(\sin x) + x \cot x) + \frac{1}{2\sqrt{x-x^2}}

Final Answer: The derivative of the given function is

(\sin x)^{x} (x \cot x + \log(\sin x)) + \frac{1}{2\sqrt{x-x^2}}

.

Q9EXERCISE 5.5

Differentiate the functions given in Exercises 1 to 11 w.r.t. $x$ . 9. $x^{\sin x}+(\sin x)^{\cos x}$

Solution

Given: The function

y = x^{\sin x}+(\sin x)^{\cos x}

.

To Find: The derivative of the function with respect to

x

.

Solution: Let the given function be

y = u + v

, where

u = x^{\sin x}

and

v = (\sin x)^{\cos x}

. Then,

\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}

.

First, consider

u = x^{\sin x}

. Using logarithmic differentiation:

\log u = \log(x^{\sin x})

\log u = (\sin x) \log x

Differentiating with respect to

x

using the product rule:

\frac{1}{u} \frac{du}{dx} = \frac{d}{dx}(\sin x) \cdot \log x + \sin x \cdot \frac{d}{dx}(\log x)

\frac{1}{u} \frac{du}{dx} = (\cos x) \log x + (\sin x) \frac{1}{x}

\frac{du}{dx} = u \left( \cos x \log x + \frac{\sin x}{x} \right)

Substituting back

u = x^{\sin x}

:

\frac{du}{dx} = x^{\sin x} \left( \cos x \log x + \frac{\sin x}{x} \right) \quad ...(1)

Next, consider

v = (\sin x)^{\cos x}

. Using logarithmic differentiation:

\log v = \log((\sin x)^{\cos x})

\log v = (\cos x) \log(\sin x)

Differentiating with respect to

x

using the product rule:

\frac{1}{v} \frac{dv}{dx} = \frac{d}{dx}(\cos x) \cdot \log(\sin x) + \cos x \cdot \frac{d}{dx}(\log(\sin x))

\frac{1}{v} \frac{dv}{dx} = (-\sin x) \log(\sin x) + \cos x \cdot \frac{1}{\sin x} \cdot \cos x

\frac{1}{v} \frac{dv}{dx} = -\sin x \log(\sin x) + \frac{\cos^2 x}{\sin x}

\frac{1}{v} \frac{dv}{dx} = -\sin x \log(\sin x) + \cot x \cos x

\frac{dv}{dx} = v (\cot x \cos x - \sin x \log(\sin x))

Substituting back

v = (\sin x)^{\cos x}

:

\frac{dv}{dx} = (\sin x)^{\cos x} (\cot x \cos x - \sin x \log(\sin x)) \quad ...(2)

Now, adding equations (1) and (2):

\frac{dy}{dx} = x^{\sin x} \left( \cos x \log x + \frac{\sin x}{x} \right) + (\sin x)^{\cos x} (\cot x \cos x - \sin x \log(\sin x))

Final Answer: The derivative of the given function is

x^{\sin x} \left( \frac{\sin x}{x} + \cos x \log x \right) + (\sin x)^{\cos x} (\cot x \cos x - \sin x \log(\sin x))

.

Q10EXERCISE 5.5

Differentiate the functions given in Exercises 1 to 11 w.r.t. $x$ . 10. $x^{x \cos x}+\frac{x^{2}+1}{x^{2}-1}$

Solution

Given: The function

y = x^{x \cos x}+\frac{x^{2}+1}{x^{2}-1}

.

To Find: The derivative of the function with respect to

x

.

Solution: Let the given function be

y = u + v

, where

u = x^{x \cos x}

and

v = \frac{x^2+1}{x^2-1}

. Then,

\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}

.

First, consider

u = x^{x \cos x}

. Using logarithmic differentiation:

\log u = \log(x^{x \cos x})

\log u = (x \cos x) \log x

Differentiating with respect to

x

using the product rule for three functions

(fgh)' = f'gh + fg'h + fgh'

:

\frac{1}{u} \frac{du}{dx} = \frac{d}{dx}(x) \cdot (\cos x \log x) + x \cdot \frac{d}{dx}(\cos x) \cdot (\log x) + (x \cos x) \cdot \frac{d}{dx}(\log x)

\frac{1}{u} \frac{du}{dx} = 1 \cdot \cos x \log x + x \cdot (-\sin x) \log x + x \cos x \cdot \frac{1}{x}

\frac{1}{u} \frac{du}{dx} = \cos x \log x - x \sin x \log x + \cos x

\frac{1}{u} \frac{du}{dx} = \cos x(1 + \log x) - x \sin x \log x

\frac{du}{dx} = u (\cos x(1 + \log x) - x \sin x \log x)

Substituting back

u = x^{x \cos x}

:

\frac{du}{dx} = x^{x \cos x} (\cos x - x \sin x \log x + \cos x \log x) \quad ...(1)

Next, consider

v = \frac{x^2+1}{x^2-1}

. Using the quotient rule,

\frac{d}{dx}\left(\frac{f}{g}\right) = \frac{f'g - fg'}{g^2}

:

\frac{dv}{dx} = \frac{\frac{d}{dx}(x^2+1) \cdot (x^2-1) - (x^2+1) \cdot \frac{d}{dx}(x^2-1)}{(x^2-1)^2}

\frac{dv}{dx} = \frac{(2x)(x^2-1) - (x^2+1)(2x)}{(x^2-1)^2}

\frac{dv}{dx} = \frac{2x^3 - 2x - 2x^3 - 2x}{(x^2-1)^2}

\frac{dv}{dx} = \frac{-4x}{(x^2-1)^2} \quad ...(2)

Now, adding equations (1) and (2):

\frac{dy}{dx} = x^{x \cos x} (\cos x - x \sin x \log x + \cos x \log x) - \frac{4x}{(x^2-1)^2}

Final Answer: The derivative of the given function is

x^{x \cos x} (\cos x(1 + \log x) - x \sin x \log x) - \frac{4x}{(x^2-1)^2}

.

Q11EXERCISE 5.5

Differentiate the functions given in Exercises 1 to 11 w.r.t. $x$ . 11. $(x \cos x)^{x}+(x \sin x)^{\frac{1}{x}}$

Solution

Given: The function

y = (x \cos x)^{x}+(x \sin x)^{\frac{1}{x}}

.

To Find: The derivative of the function with respect to

x

.

Solution: Let the given function be

y = u + v

, where

u = (x \cos x)^{x}

and

v = (x \sin x)^{\frac{1}{x}}

. Then,

\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}

.

First, consider

u = (x \cos x)^{x}

. Using logarithmic differentiation:

\log u = \log((x \cos x)^{x})

\log u = x \log(x \cos x) = x(\log x + \log(\cos x))

Differentiating with respect to

x

using the product rule:

\frac{1}{u} \frac{du}{dx} = \frac{d}{dx}(x) \cdot (\log x + \log(\cos x)) + x \cdot \frac{d}{dx}(\log x + \log(\cos x))

\frac{1}{u} \frac{du}{dx} = 1 \cdot (\log x + \log(\cos x)) + x \left( \frac{1}{x} + \frac{1}{\cos x}(-\sin x) \right)

\frac{1}{u} \frac{du}{dx} = \log(x \cos x) + x \left( \frac{1}{x} - \tan x \right)

\frac{1}{u} \frac{du}{dx} = \log(x \cos x) + 1 - x \tan x

\frac{du}{dx} = u (1 - x \tan x + \log(x \cos x))

Substituting back

u = (x \cos x)^{x}

:

\frac{du}{dx} = (x \cos x)^{x} (1 - x \tan x + \log(x \cos x)) \quad ...(1)

Next, consider

v = (x \sin x)^{\frac{1}{x}}

. Using logarithmic differentiation:

\log v = \log((x \sin x)^{\frac{1}{x}})

\log v = \frac{1}{x} \log(x \sin x) = \frac{\log x + \log(\sin x)}{x}

Differentiating with respect to

x

using the quotient rule:

\frac{1}{v} \frac{dv}{dx} = \frac{\frac{d}{dx}(\log x + \log(\sin x)) \cdot x - (\log x + \log(\sin x)) \cdot \frac{d}{dx}(x)}{x^2}

\frac{1}{v} \frac{dv}{dx} = \frac{\left(\frac{1}{x} + \frac{\cos x}{\sin x}\right)x - (\log(x \sin x)) \cdot 1}{x^2}

\frac{1}{v} \frac{dv}{dx} = \frac{(1 + x \cot x) - \log(x \sin x)}{x^2}

\frac{dv}{dx} = v \left( \frac{1 + x \cot x - \log(x \sin x)}{x^2} \right)

Substituting back

v = (x \sin x)^{\frac{1}{x}}

:

\frac{dv}{dx} = (x \sin x)^{\frac{1}{x}} \left( \frac{1 + x \cot x - \log(x \sin x)}{x^2} \right) \quad ...(2)

Now, adding equations (1) and (2):

\frac{dy}{dx} = (x \cos x)^{x} (1 - x \tan x + \log(x \cos x)) + (x \sin x)^{\frac{1}{x}} \left( \frac{1 + x \cot x - \log(x \sin x)}{x^2} \right)

Final Answer: The derivative of the given function is

(x \cos x)^{x} [1 - x \tan x + \log(x \cos x)] + (x \sin x)^{\frac{1}{x}} [\frac{x \cot x + 1 - \log(x \sin x)}{x^2}]

.

Q12EXERCISE 5.5

Find $\frac{d y}{d x}$ of the functions given in Exercises 12 to 15. 12. $x^{y}+y^{x}=1$

Solution

Given: The function is

x^{y} + y^{x} = 1

.

To Find:

\frac{dy}{dx}

Solution: Let

u = x^y

and

v = y^x

. The equation becomes

u + v = 1

. Differentiating with respect to

x

, we get:

\frac{du}{dx} + \frac{dv}{dx} = \frac{d(1)}{dx}

\frac{du}{dx} + \frac{dv}{dx} = 0 \quad ...(1)

First, consider

u = x^y

. Taking logarithm on both sides:

\log u = \log(x^y)

\log u = y \log x

Differentiating with respect to

x

using the product rule:

\frac{1}{u} \frac{du}{dx} = \frac{dy}{dx} \cdot \log x + y \cdot \frac{1}{x}

\frac{du}{dx} = u \left( \frac{dy}{dx} \log x + \frac{y}{x} \right)

Substituting

u = x^y

:

\frac{du}{dx} = x^y \left( \log x \frac{dy}{dx} + \frac{y}{x} \right) \quad ...(2)

Next, consider

v = y^x

. Taking logarithm on both sides:

\log v = \log(y^x)

\log v = x \log y

Differentiating with respect to

x

using the product rule:

\frac{1}{v} \frac{dv}{dx} = 1 \cdot \log y + x \cdot \frac{1}{y} \frac{dy}{dx}

\frac{dv}{dx} = v \left( \log y + \frac{x}{y} \frac{dy}{dx} \right)

Substituting

v = y^x

:

\frac{dv}{dx} = y^x \left( \log y + \frac{x}{y} \frac{dy}{dx} \right) \quad ...(3)

Now, substitute equations (2) and (3) into equation (1):

x^y \left( \log x \frac{dy}{dx} + \frac{y}{x} \right) + y^x \left( \log y + \frac{x}{y} \frac{dy}{dx} \right) = 0

x^y \log x \frac{dy}{dx} + x^y \frac{y}{x} + y^x \log y + y^x \frac{x}{y} \frac{dy}{dx} = 0

x^y \log x \frac{dy}{dx} + y x^{y-1} + y^x \log y + x y^{x-1} \frac{dy}{dx} = 0

Group the terms with

\frac{dy}{dx}

:

\frac{dy}{dx} (x^y \log x + x y^{x-1}) = - (y x^{y-1} + y^x \log y)

\frac{dy}{dx} = - \frac{y x^{y-1} + y^x \log y}{x y^{x-1} + x^y \log x}

Final Answer:

\frac{dy}{dx} = - \frac{y^x \log y + y x^{y-1}}{x^y \log x + x y^{x-1}}

Q13EXERCISE 5.5

Find $\frac{d y}{d x}$ of the functions given in Exercises 12 to 15. 13. $y^{x}=x^{y}$

Solution

Given: The function is

y^x = x^y

.

To Find:

\frac{dy}{dx}

Solution: The given equation is

y^x = x^y

. Since the function involves variables in both the base and the exponent, we use logarithmic differentiation. Taking logarithm on both sides:

\log(y^x) = \log(x^y)

x \log y = y \log x

Differentiating both sides with respect to

x

using the product rule:

\frac{d}{dx}(x \log y) = \frac{d}{dx}(y \log x)

(1 \cdot \log y + x \cdot \frac{1}{y} \frac{dy}{dx}) = (\frac{dy}{dx} \cdot \log x + y \cdot \frac{1}{x})

\log y + \frac{x}{y} \frac{dy}{dx} = \log x \frac{dy}{dx} + \frac{y}{x}

Now, we rearrange the terms to solve for

\frac{dy}{dx}

.

\frac{x}{y} \frac{dy}{dx} - \log x \frac{dy}{dx} = \frac{y}{x} - \log y

Factor out

\frac{dy}{dx}

on the left side:

\frac{dy}{dx} \left( \frac{x}{y} - \log x \right) = \frac{y}{x} - \log y

Simplify the expressions inside the brackets:

\frac{dy}{dx} \left( \frac{x - y \log x}{y} \right) = \frac{y - x \log y}{x}

Isolate

\frac{dy}{dx}

:

\frac{dy}{dx} = \frac{y - x \log y}{x} \cdot \frac{y}{x - y \log x}

\frac{dy}{dx} = \frac{y(y - x \log y)}{x(x - y \log x)}

Final Answer:

\frac{dy}{dx} = \frac{y(y - x \log y)}{x(x - y \log x)}

Q14EXERCISE 5.5

Find $\frac{d y}{d x}$ of the functions given in Exercises 12 to 15. 14. $(\cos x)^{y}=(\cos y)^{x}$

Solution

Given: The function is

(\cos x)^{y} = (\cos y)^{x}

.

To Find:

\frac{dy}{dx}

Solution: The given equation is

(\cos x)^{y} = (\cos y)^{x}

. We use logarithmic differentiation because variables are present in both the base and the exponent. Taking logarithm on both sides:

\log((\cos x)^y) = \log((\cos y)^x)

y \log(\cos x) = x \log(\cos y)

Differentiating both sides with respect to

x

using the product rule and chain rule:

\frac{d}{dx}(y \log(\cos x)) = \frac{d}{dx}(x \log(\cos y))

\frac{dy}{dx} \cdot \log(\cos x) + y \cdot \frac{1}{\cos x} \cdot (-\sin x) = 1 \cdot \log(\cos y) + x \cdot \frac{1}{\cos y} \cdot (-\sin y) \cdot \frac{dy}{dx}

\frac{dy}{dx} \log(\cos x) - y \tan x = \log(\cos y) - x \tan y \frac{dy}{dx}

Now, we group the terms containing

\frac{dy}{dx}

on one side:

\frac{dy}{dx} \log(\cos x) + x \tan y \frac{dy}{dx} = \log(\cos y) + y \tan x

Factor out

\frac{dy}{dx}

:

\frac{dy}{dx} (\log(\cos x) + x \tan y) = \log(\cos y) + y \tan x

Isolate

\frac{dy}{dx}

:

\frac{dy}{dx} = \frac{\log(\cos y) + y \tan x}{\log(\cos x) + x \tan y}

Final Answer:

\frac{dy}{dx} = \frac{\log(\cos y) + y \tan x}{\log(\cos x) + x \tan y}

Q15EXERCISE 5.5

Find $\frac{d y}{d x}$ of the functions given in Exercises 12 to 15. 15. $x y=e^{(x-y)}$

Solution

Given: The function is

xy = e^{(x-y)}

.

To Find:

\frac{dy}{dx}

Solution: The given equation is

xy = e^{(x-y)}

. Taking logarithm on both sides is an efficient method to solve this.

\log(xy) = \log(e^{x-y})

Using logarithm properties,

\log(ab) = \log a + \log b

and

\log(e^z) = z

:

\log x + \log y = x - y

Now, differentiate both sides with respect to

x

:

\frac{d}{dx}(\log x + \log y) = \frac{d}{dx}(x - y)

\frac{1}{x} + \frac{1}{y} \frac{dy}{dx} = 1 - \frac{dy}{dx}

Rearrange the terms to group

\frac{dy}{dx}

terms together:

\frac{1}{y} \frac{dy}{dx} + \frac{dy}{dx} = 1 - \frac{1}{x}

Factor out

\frac{dy}{dx}

:

\frac{dy}{dx} \left( \frac{1}{y} + 1 \right) = 1 - \frac{1}{x}

Simplify the expressions in the parentheses:

\frac{dy}{dx} \left( \frac{1+y}{y} \right) = \frac{x-1}{x}

Now, solve for

\frac{dy}{dx}

:

\frac{dy}{dx} = \frac{x-1}{x} \cdot \frac{y}{1+y}

\frac{dy}{dx} = \frac{y(x-1)}{x(1+y)}

Final Answer:

\frac{dy}{dx} = \frac{y(x-1)}{x(y+1)}

Q16EXERCISE 5.5

Find the derivative of the function given by $f(x)=(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right)\left(1+x^{8}\right)$ and hence find $f'(1)$ .

Solution

Given: The function

f(x) = (1+x)(1+x^2)(1+x^4)(1+x^8)

.

To Find: The derivative

f'(x)

and the value of

f'(1)

.

Solution: Let

y = f(x) = (1+x)(1+x^2)(1+x^4)(1+x^8)

. Since the function is a product of several terms, we use logarithmic differentiation. Taking logarithm on both sides:

\log y = \log[(1+x)(1+x^2)(1+x^4)(1+x^8)]

Using the property

\log(abcd) = \log a + \log b + \log c + \log d

:

\log y = \log(1+x) + \log(1+x^2) + \log(1+x^4) + \log(1+x^8)

Differentiating both sides with respect to

x

:

\frac{d}{dx}(\log y) = \frac{d}{dx}[\log(1+x) + \log(1+x^2) + \log(1+x^4) + \log(1+x^8)]

\frac{1}{y} \frac{dy}{dx} = \frac{1}{1+x} \cdot 1 + \frac{1}{1+x^2} \cdot 2x + \frac{1}{1+x^4} \cdot 4x^3 + \frac{1}{1+x^8} \cdot 8x^7

\frac{dy}{dx} = y \left[ \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \frac{8x^7}{1+x^8} \right]

Substituting

y = f(x)

back, we get the derivative

f'(x)

:

f'(x) = (1+x)(1+x^2)(1+x^4)(1+x^8) \left[ \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \frac{8x^7}{1+x^8} \right]

Now, we need to find

f'(1)

. We substitute

x=1

into the expression for

f'(x)

. First, let's calculate the value of the pre-factor at

x=1

:

f(1) = (1+1)(1+1^2)(1+1^4)(1+1^8) = (2)(2)(2)(2) = 16

Now, calculate the value of the expression in the brackets at

x=1

:

\left[ \frac{1}{1+1} + \frac{2(1)}{1+1^2} + \frac{4(1)^3}{1+1^4} + \frac{8(1)^7}{1+1^8} \right] = \left[ \frac{1}{2} + \frac{2}{2} + \frac{4}{2} + \frac{8}{2} \right]

= \frac{1}{2} + 1 + 2 + 4 = 7 + \frac{1}{2} = \frac{14+1}{2} = \frac{15}{2}

Therefore,

f'(1)

is the product of these two values:

f'(1) = f(1) \times \left( \frac{15}{2} \right)

f'(1) = 16 \times \frac{15}{2}

f'(1) = 8 \times 15 = 120

Final Answer: The derivative of the function is

f'(x) = (1+x)(1+x^2)(1+x^4)(1+x^8) \left[ \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \frac{8x^7}{1+x^8} \right]

. The value of

f'(1)

is

120

.

Q17EXERCISE 5.5

Differentiate $\left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right)$ in three ways mentioned below:

(i)

by using product rule

(ii)

by expanding the product to obtain a single polynomial.

(iii)

by logarithmic differentiation. Do they all give the same answer?

Solution

Given: The function to differentiate is

f(x) = \left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right)

.

To Find: The derivative of the function in three different ways and check if the answers are the same.

Let

y = \left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right)

.

(i) By using product rule Let

u(x) = x^{2}-5 x+8

and

v(x) = x^{3}+7 x+9

. Then

y = u(x) \cdot v(x)

. The product rule states that

\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}

.

First, find the derivatives of

u

and

v

:

\frac{du}{dx} = \frac{d}{dx}(x^{2}-5 x+8) = 2x - 5

\frac{dv}{dx} = \frac{d}{dx}(x^{3}+7 x+9) = 3x^2 + 7

Now, apply the product rule:

\frac{dy}{dx} = (x^{2}-5 x+8)(3x^2 + 7) + (x^{3}+7 x+9)(2x - 5)

Expand the products:

= (x^2(3x^2+7) - 5x(3x^2+7) + 8(3x^2+7)) + (x^3(2x-5) + 7x(2x-5) + 9(2x-5))

= (3x^4+7x^2 - 15x^3-35x + 24x^2+56) + (2x^4-5x^3 + 14x^2-35x + 18x-45)

= (3x^4 - 15x^3 + 31x^2 - 35x + 56) + (2x^4 - 5x^3 + 14x^2 - 17x - 45)

Combine like terms:

= (3+2)x^4 + (-15-5)x^3 + (31+14)x^2 + (-35-17)x + (56-45)

= 5x^4 - 20x^3 + 45x^2 - 52x + 11

(ii) By expanding the product to obtain a single polynomial First, expand the product

y = \left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right)

.

y = x^2(x^3+7x+9) - 5x(x^3+7x+9) + 8(x^3+7x+9)

y = (x^5+7x^3+9x^2) - (5x^4+35x^2+45x) + (8x^3+56x+72)

Combine like terms:

y = x^5 - 5x^4 + (7+8)x^3 + (9-35)x^2 + (-45+56)x + 72

y = x^5 - 5x^4 + 15x^3 - 26x^2 + 11x + 72

Now, differentiate this polynomial with respect to

x

:

\frac{dy}{dx} = \frac{d}{dx}(x^5 - 5x^4 + 15x^3 - 26x^2 + 11x + 72)

\frac{dy}{dx} = 5x^{5-1} - 5(4x^{4-1}) + 15(3x^{3-1}) - 26(2x^{2-1}) + 11(1) + 0

\frac{dy}{dx} = 5x^4 - 20x^3 + 45x^2 - 52x + 11

(iii) By logarithmic differentiation Let

y = \left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right)

. Take the natural logarithm of both sides:

\ln y = \ln \left[ \left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right) \right]

Using the logarithm property

\ln(ab) = \ln a + \ln b

:

\ln y = \ln(x^{2}-5 x+8) + \ln(x^{3}+7 x+9)

Differentiate both sides with respect to

x

:

\frac{1}{y} \frac{dy}{dx} = \frac{1}{x^{2}-5 x+8} \cdot \frac{d}{dx}(x^{2}-5 x+8) + \frac{1}{x^{3}+7 x+9} \cdot \frac{d}{dx}(x^{3}+7 x+9)

\frac{1}{y} \frac{dy}{dx} = \frac{2x-5}{x^{2}-5 x+8} + \frac{3x^2+7}{x^{3}+7 x+9}

Multiply by

y

to solve for

\frac{dy}{dx}

:

\frac{dy}{dx} = y \left[ \frac{2x-5}{x^{2}-5 x+8} + \frac{3x^2+7}{x^{3}+7 x+9} \right]

Substitute back

y = \left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right)

:

\frac{dy}{dx} = \left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right) \left[ \frac{2x-5}{x^{2}-5 x+8} + \frac{3x^2+7}{x^{3}+7 x+9} \right]

Distribute the term outside the brackets:

\frac{dy}{dx} = (x^{3}+7 x+9)(2x-5) + (x^{2}-5 x+8)(3x^2+7)

This is the same expression obtained in the product rule method. Expanding it gives:

\frac{dy}{dx} = 5x^4 - 20x^3 + 45x^2 - 52x + 11

Conclusion: Yes, all three methods give the same answer.

Final Answer: The derivative is

5x^4 - 20x^3 + 45x^2 - 52x + 11

. All three methods yield the same result.

Q18EXERCISE 5.5

If $u, v$ and $w$ are functions of $x$ , then show that $\frac{d}{d x}(u . v . w)=\frac{d u}{d x} v . w+u \cdot \frac{d v}{d x} \cdot w+u \cdot v \frac{d w}{d x}$ in two ways - first by repeated application of product rule, second by logarithmic differentiation.

Solution

To Prove: If

u, v,

and

w

are functions of

x

, then

\frac{d}{d x}(u . v . w)=\frac{d u}{d x} v . w+u . \frac{d v}{d x} . w+u . v \frac{d w}{d x}

Proof: Let

y = u \cdot v \cdot w

.

Method 1: By repeated application of the product rule

Let's group the terms as

y = (u \cdot v) \cdot w

. We can apply the product rule by treating

(u \cdot v)

as the first function and

w

as the second function.

Using the product rule

\frac{d}{dx}(f \cdot g) = f \frac{dg}{dx} + g \frac{df}{dx}

:

\frac{dy}{dx} = \frac{d}{dx}[(u \cdot v) \cdot w] = (u \cdot v) \frac{dw}{dx} + w \frac{d}{dx}(u \cdot v) \quad ... (1)

Now, we need to find

\frac{d}{dx}(u \cdot v)

. We apply the product rule again:

\frac{d}{dx}(u \cdot v) = u \frac{dv}{dx} + v \frac{du}{dx} \quad ... (2)

Substitute equation (2) into equation (1):

\frac{dy}{dx} = (u \cdot v) \frac{dw}{dx} + w \left( u \frac{dv}{dx} + v \frac{du}{dx} \right)

Distribute

w

into the parenthesis:

\frac{dy}{dx} = u \cdot v \frac{dw}{dx} + u \cdot w \frac{dv}{dx} + v \cdot w \frac{du}{dx}

Rearranging the terms to match the desired format:

\frac{d}{dx}(u \cdot v \cdot w) = \frac{du}{dx} v \cdot w + u \frac{dv}{dx} w + u \cdot v \frac{dw}{dx}

Hence Proved using the product rule.

Method 2: By logarithmic differentiation

Let

y = u \cdot v \cdot w

. Take the natural logarithm of both sides:

\ln y = \ln(u \cdot v \cdot w)

Using the property of logarithms,

\ln(abc) = \ln a + \ln b + \ln c

:

\ln y = \ln u + \ln v + \ln w

Differentiate both sides with respect to

x

:

\frac{d}{dx}(\ln y) = \frac{d}{dx}(\ln u) + \frac{d}{dx}(\ln v) + \frac{d}{dx}(\ln w)

Using the chain rule,

\frac{d}{dx}(\ln f(x)) = \frac{1}{f(x)} f'(x)

:

\frac{1}{y} \frac{dy}{dx} = \frac{1}{u} \frac{du}{dx} + \frac{1}{v} \frac{dv}{dx} + \frac{1}{w} \frac{dw}{dx}

To isolate

\frac{dy}{dx}

, multiply both sides by

y

:

\frac{dy}{dx} = y \left( \frac{1}{u} \frac{du}{dx} + \frac{1}{v} \frac{dv}{dx} + \frac{1}{w} \frac{dw}{dx} \right)

Substitute back

y = u \cdot v \cdot w

:

\frac{dy}{dx} = (u \cdot v \cdot w) \left( \frac{1}{u} \frac{du}{dx} + \frac{1}{v} \frac{dv}{dx} + \frac{1}{w} \frac{dw}{dx} \right)

Distribute the

(u \cdot v \cdot w)

term:

\frac{dy}{dx} = \frac{u \cdot v \cdot w}{u} \frac{du}{dx} + \frac{u \cdot v \cdot w}{v} \frac{dv}{dx} + \frac{u \cdot v \cdot w}{w} \frac{dw}{dx}

Simplify by cancelling terms:

\frac{dy}{dx} = (v \cdot w) \frac{du}{dx} + (u \cdot w) \frac{dv}{dx} + (u \cdot v) \frac{dw}{dx}

Rearranging the terms:

\frac{d}{dx}(u \cdot v \cdot w) = \frac{du}{dx} v \cdot w + u \frac{dv}{dx} w + u \cdot v \frac{dw}{dx}

Hence Proved using logarithmic differentiation.

Q1EXERCISE 5.6

If $x$ and $y$ are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find $\frac{d y}{d x}$ . $x=2 a t^{2}, y=a t^{4}$

Solution

Given: The parametric equations are:

x = 2at^2

y = at^4

To Find:

\frac{dy}{dx}

without eliminating the parameter

t

.

Formula: For parametric functions

x = f(t)

and

y = g(t)

, the derivative is given by:

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}

Solution: First, we differentiate

x

with respect to the parameter

t

.

x = 2at^2

\frac{dx}{dt} = \frac{d}{dt}(2at^2) = 2a \cdot (2t) = 4at

Next, we differentiate

y

with respect to the parameter

t

.

y = at^4

\frac{dy}{dt} = \frac{d}{dt}(at^4) = a \cdot (4t^3) = 4at^3

Now, we can find

\frac{dy}{dx}

using the formula for parametric differentiation:

\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4at^3}{4at}

Assuming

a \neq 0

and

t \neq 0

, we can cancel the common terms

4at

:

\frac{dy}{dx} = t^2

Final Answer: The value of

\frac{dy}{dx}

is

t^2

.

Q2EXERCISE 5.6

$x=a \cos \theta, y=b \cos \theta$

Solution

Given: The parametric equations are:

x = a \cos \theta

y = b \cos \theta

To Find:

\frac{dy}{dx}

without eliminating the parameter

\theta

.

Formula: For parametric functions

x = f(\theta)

and

y = g(\theta)

, the derivative is given by:

\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}

Solution: First, we differentiate

x

with respect to the parameter

\theta

.

x = a \cos \theta

\frac{dx}{d\theta} = \frac{d}{d\theta}(a \cos \theta) = a(-\sin \theta) = -a \sin \theta

Next, we differentiate

y

with respect to the parameter

\theta

.

y = b \cos \theta

\frac{dy}{d\theta} = \frac{d}{d\theta}(b \cos \theta) = b(-\sin \theta) = -b \sin \theta

Now, we can find

\frac{dy}{dx}

using the formula for parametric differentiation:

\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{-b \sin \theta}{-a \sin \theta}

Assuming

\sin \theta \neq 0

, we can cancel the common term

-\sin \theta

:

\frac{dy}{dx} = \frac{b}{a}

Final Answer: The value of

\frac{dy}{dx}

is

\frac{b}{a}

.

Q3EXERCISE 5.6

$x=\sin t, y=\cos 2 t$

Solution

Given: The parametric equations are:

x = \sin t

y = \cos 2t

To Find:

\frac{dy}{dx}

without eliminating the parameter

t

.

Formula: For parametric functions

x = f(t)

and

y = g(t)

, the derivative is given by:

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}

Solution: First, we differentiate

x

with respect to the parameter

t

.

x = \sin t

\frac{dx}{dt} = \frac{d}{dt}(\sin t) = \cos t

Next, we differentiate

y

with respect to the parameter

t

.

y = \cos 2t

Using the chain rule:

\frac{dy}{dt} = \frac{d}{dt}(\cos 2t) = -\sin(2t) \cdot \frac{d}{dt}(2t) = -2\sin(2t)

Now, we can find

\frac{dy}{dx}

using the formula for parametric differentiation:

\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-2\sin(2t)}{\cos t}

To simplify the expression, we use the double angle identity for sine:

\sin(2t) = 2\sin t \cos t

.

\frac{dy}{dx} = \frac{-2(2\sin t \cos t)}{\cos t}

Assuming

\cos t \neq 0

, we can cancel the common term

\cos t

:

\frac{dy}{dx} = -4\sin t

Final Answer: The value of

\frac{dy}{dx}

is

-4\sin t

.

Q4EXERCISE 5.6

$x=4 t, y=\frac{4}{t}$

Solution

Given: The parametric equations are:

x = 4t

y = \frac{4}{t}

To Find: The derivative

\frac{dy}{dx}

.

Solution: We have the parametric equations

x = x(t)

and

y = y(t)

. The derivative

\frac{dy}{dx}

is given by the formula:

\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}

First, we differentiate

x

with respect to

t

:

x = 4t

\frac{dx}{dt} = \frac{d}{dt}(4t) = 4

Next, we differentiate

y

with respect to

t

:

y = \frac{4}{t} = 4t^{-1}

\frac{dy}{dt} = \frac{d}{dt}(4t^{-1}) = 4(-1)t^{-1-1} = -4t^{-2} = -\frac{4}{t^2}

Now, we can find

\frac{dy}{dx}

:

\frac{dy}{dx} = \frac{-4/t^2}{4} = -\frac{4}{t^2} \times \frac{1}{4} = -\frac{1}{t^2}

Final Answer: The derivative

\frac{dy}{dx}

is

-\frac{1}{t^2}

.

Q5EXERCISE 5.6

$x=\cos \theta-\cos 2 \theta, y=\sin \theta-\sin 2 \theta$

Solution

Given: The parametric equations are:

x = \cos \theta - \cos 2\theta

y = \sin \theta - \sin 2\theta

To Find: The derivative

\frac{dy}{dx}

.

Solution: We have the parametric equations where

x

and

y

are functions of the parameter

\theta

. The derivative

\frac{dy}{dx}

is given by the formula:

\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}

First, we differentiate

x

with respect to

\theta

:

x = \cos \theta - \cos 2\theta

\frac{dx}{d\theta} = \frac{d}{d\theta}(\cos \theta - \cos 2\theta)

= -\sin \theta - (-\sin 2\theta) \cdot \frac{d}{d\theta}(2\theta)

= -\sin \theta + 2\sin 2\theta

Next, we differentiate

y

with respect to

\theta

:

y = \sin \theta - \sin 2\theta

\frac{dy}{d\theta} = \frac{d}{d\theta}(\sin \theta - \sin 2\theta)

= \cos \theta - (\cos 2\theta) \cdot \frac{d}{d\theta}(2\theta)

= \cos \theta - 2\cos 2\theta

Now, we can find

\frac{dy}{dx}

:

\frac{dy}{dx} = \frac{\cos \theta - 2\cos 2\theta}{2\sin 2\theta - \sin \theta}

Final Answer: The derivative

\frac{dy}{dx}

is

\frac{\cos \theta - 2\cos 2\theta}{2\sin 2\theta - \sin \theta}

.

Q6EXERCISE 5.6

$x=a(\theta-\sin \theta), y=a(1+\cos \theta)$

Solution

Given: The parametric equations are:

x = a(\theta - \sin \theta)

y = a(1 + \cos \theta)

To Find: The derivative

\frac{dy}{dx}

.

Solution: We have the parametric equations where

x

and

y

are functions of the parameter

\theta

. The derivative

\frac{dy}{dx}

is given by the formula:

\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}

First, we differentiate

x

with respect to

\theta

:

x = a(\theta - \sin \theta)

\frac{dx}{d\theta} = a \frac{d}{d\theta}(\theta - \sin \theta) = a(1 - \cos \theta)

Next, we differentiate

y

with respect to

\theta

:

y = a(1 + \cos \theta)

\frac{dy}{d\theta} = a \frac{d}{d\theta}(1 + \cos \theta) = a(0 - \sin \theta) = -a\sin \theta

Now, we can find

\frac{dy}{dx}

:

\frac{dy}{dx} = \frac{-a\sin \theta}{a(1 - \cos \theta)} = -\frac{\sin \theta}{1 - \cos \theta}

To simplify the expression, we use the half-angle trigonometric identities:

\sin \theta = 2\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})

1 - \cos \theta = 2\sin^2(\frac{\theta}{2})

Substituting these into our expression for

\frac{dy}{dx}

:

\frac{dy}{dx} = -\frac{2\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})}{2\sin^2(\frac{\theta}{2})}

\frac{dy}{dx} = -\frac{\cos(\frac{\theta}{2})}{\sin(\frac{\theta}{2})} = -\cot(\frac{\theta}{2})

Final Answer: The derivative

\frac{dy}{dx}

is

-\cot(\frac{\theta}{2})

.

Q7EXERCISE 5.6

$x=\frac{\sin ^{3} t}{\sqrt{\cos 2 t}}, y=\frac{\cos ^{3} t}{\sqrt{\cos 2 t}}$

Solution

Given: The parametric equations are:

x = \frac{\sin^3 t}{\sqrt{\cos 2t}}

y = \frac{\cos^3 t}{\sqrt{\cos 2t}}

To Find: The derivative

\frac{dy}{dx}

.

Solution: We have the parametric equations where

x

and

y

are functions of the parameter

t

. The derivative

\frac{dy}{dx}

is given by the formula:

\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}

We will use logarithmic differentiation to find

\frac{dx}{dt}

and

\frac{dy}{dt}

.

For

x = \frac{\sin^3 t}{\sqrt{\cos 2t}}

: Taking logarithm on both sides:

\log x = \log(\sin^3 t) - \log(\sqrt{\cos 2t})

\log x = 3\log(\sin t) - \frac{1}{2}\log(\cos 2t)

Differentiating with respect to

t

:

\frac{1}{x}\frac{dx}{dt} = 3\frac{\cos t}{\sin t} - \frac{1}{2}\frac{-2\sin 2t}{\cos 2t} = 3\cot t + \tan 2t

\frac{dx}{dt} = x(3\cot t + \tan 2t) = \frac{\sin^3 t}{\sqrt{\cos 2t}} \left( \frac{3\cos t}{\sin t} + \frac{\sin 2t}{\cos 2t} \right)

\frac{dx}{dt} = \frac{\sin^3 t}{\sqrt{\cos 2t}} \left( \frac{3\cos t \cos 2t + \sin t \sin 2t}{\sin t \cos 2t} \right)

\frac{dx}{dt} = \frac{\sin^2 t}{(\cos 2t)^{3/2}} (2\cos t \cos 2t + \cos t \cos 2t + \sin t \sin 2t)

Using

\cos(A-B) = \cos A \cos B + \sin A \sin B

:

\frac{dx}{dt} = \frac{\sin^2 t}{(\cos 2t)^{3/2}} (2\cos t \cos 2t + \cos(2t-t))

\frac{dx}{dt} = \frac{\sin^2 t \cos t (2\cos 2t + 1)}{(\cos 2t)^{3/2}}

For

y = \frac{\cos^3 t}{\sqrt{\cos 2t}}

: Taking logarithm on both sides:

\log y = 3\log(\cos t) - \frac{1}{2}\log(\cos 2t)

Differentiating with respect to

t

:

\frac{1}{y}\frac{dy}{dt} = 3\frac{-\sin t}{\cos t} - \frac{1}{2}\frac{-2\sin 2t}{\cos 2t} = -3\tan t + \tan 2t

\frac{dy}{dt} = y(-3\tan t + \tan 2t) = \frac{\cos^3 t}{\sqrt{\cos 2t}} \left( -\frac{3\sin t}{\cos t} + \frac{\sin 2t}{\cos 2t} \right)

\frac{dy}{dt} = \frac{\cos^3 t}{\sqrt{\cos 2t}} \left( \frac{-3\sin t \cos 2t + \cos t \sin 2t}{\cos t \cos 2t} \right)

\frac{dy}{dt} = \frac{\cos^2 t}{(\cos 2t)^{3/2}} (-3\sin t \cos 2t + \sin 2t \cos t)

Using

\sin(A-B) = \sin A \cos B - \cos A \sin B

:

\frac{dy}{dt} = \frac{\cos^2 t}{(\cos 2t)^{3/2}} (-2\sin t \cos 2t - \sin t \cos 2t + \sin 2t \cos t)

\frac{dy}{dt} = \frac{\cos^2 t}{(\cos 2t)^{3/2}} (-2\sin t \cos 2t + \sin(2t-t))

\frac{dy}{dt} = \frac{\cos^2 t \sin t (-2\cos 2t + 1)}{(\cos 2t)^{3/2}}

Now, we find

\frac{dy}{dx}

:

\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{\cos^2 t \sin t (1 - 2\cos 2t)}{(\cos 2t)^{3/2}}}{\frac{\sin^2 t \cos t (1 + 2\cos 2t)}{(\cos 2t)^{3/2}}}

\frac{dy}{dx} = \frac{\cos t (1 - 2\cos 2t)}{\sin t (1 + 2\cos 2t)}

Using

\cos 2t = 2\cos^2 t - 1

and

\cos 2t = 1 - 2\sin^2 t

: Numerator:

\cos t (1 - 2(2\cos^2 t - 1)) = \cos t (3 - 4\cos^2 t) = - (4\cos^3 t - 3\cos t) = -\cos 3t

Denominator:

\sin t (1 + 2(1 - 2\sin^2 t)) = \sin t (3 - 4\sin^2 t) = 3\sin t - 4\sin^3 t = \sin 3t

Therefore,

\frac{dy}{dx} = \frac{-\cos 3t}{\sin 3t} = -\cot 3t

Final Answer: The derivative

\frac{dy}{dx}

is

-\cot 3t

.

Q8EXERCISE 5.6

$x=a\left(\cos t+\log \tan \frac{t}{2}\right) y=a \sin t$

Solution

Given: The parametric equations are:

x = a\left(\cos t + \log \tan \frac{t}{2}\right)

y = a \sin t

To Find: The derivative

\frac{dy}{dx}

.

Solution: We have the parametric equations where

x

and

y

are functions of the parameter

t

. The derivative

\frac{dy}{dx}

is given by the formula:

\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}

First, we differentiate

y

with respect to

t

:

y = a \sin t

\frac{dy}{dt} = a \cos t

Next, we differentiate

x

with respect to

t

:

x = a\left(\cos t + \log \tan \frac{t}{2}\right)

\frac{dx}{dt} = a \frac{d}{dt}\left(\cos t + \log \tan \frac{t}{2}\right)

= a \left[ -\sin t + \frac{d}{dt}\left(\log \tan \frac{t}{2}\right) \right]

Using the chain rule for the log term:

\frac{d}{dt}\left(\log \tan \frac{t}{2}\right) = \frac{1}{\tan \frac{t}{2}} \cdot \frac{d}{dt}\left(\tan \frac{t}{2}\right)

= \frac{1}{\tan \frac{t}{2}} \cdot \sec^2\left(\frac{t}{2}\right) \cdot \frac{d}{dt}\left(\frac{t}{2}\right)

= \frac{\cos(t/2)}{\sin(t/2)} \cdot \frac{1}{\cos^2(t/2)} \cdot \frac{1}{2}

= \frac{1}{2\sin(t/2)\cos(t/2)}

Using the double angle identity

\sin(2A) = 2\sin A \cos A

, we have

2\sin(t/2)\cos(t/2) = \sin t

.

= \frac{1}{\sin t}

Substituting this back into the expression for

\frac{dx}{dt}

:

\frac{dx}{dt} = a \left( -\sin t + \frac{1}{\sin t} \right)

= a \left( \frac{-\sin^2 t + 1}{\sin t} \right)

Using the identity

\sin^2 t + \cos^2 t = 1

, we get

1 - \sin^2 t = \cos^2 t

.

\frac{dx}{dt} = a \frac{\cos^2 t}{\sin t}

Now, we can find

\frac{dy}{dx}

:

\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{a \cos t}{a \frac{\cos^2 t}{\sin t}}

\frac{dy}{dx} = \frac{a \cos t}{1} \cdot \frac{\sin t}{a \cos^2 t}

\frac{dy}{dx} = \frac{\sin t}{\cos t} = \tan t

Final Answer: The derivative

\frac{dy}{dx}

is

\tan t

.

Q9EXERCISE 5.6

$x=a \sec \theta, y=b \tan \theta$

Solution

Given: The parametric equations for a curve are:

x = a \sec \theta

y = b \tan \theta

To Find: The derivative

\frac{dy}{dx}

.

Solution: This is a parametric differentiation problem. We will find

\frac{dx}{d\theta}

and

\frac{dy}{d\theta}

first, and then use the chain rule

\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}

.

Step 1: Differentiate

x

with respect to

\theta

.

x = a \sec \theta

\frac{dx}{d\theta} = \frac{d}{d\theta}(a \sec \theta) = a \frac{d}{d\theta}(\sec \theta)

\frac{dx}{d\theta} = a \sec \theta \tan \theta

Step 2: Differentiate

y

with respect to

\theta

.

y = b \tan \theta

\frac{dy}{d\theta} = \frac{d}{d\theta}(b \tan \theta) = b \frac{d}{d\theta}(\tan \theta)

\frac{dy}{d\theta} = b \sec^2 \theta

Step 3: Find

\frac{dy}{dx}

.

\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{b \sec^2 \theta}{a \sec \theta \tan \theta}

Simplifying the expression:

\frac{dy}{dx} = \frac{b \sec \theta}{a \tan \theta}

Now, we express

\sec \theta

and

\tan \theta

in terms of

\sin \theta

and

\cos \theta

.

\frac{dy}{dx} = \frac{b}{a} \cdot \frac{\frac{1}{\cos \theta}}{\frac{\sin \theta}{\cos \theta}} = \frac{b}{a} \cdot \frac{1}{\sin \theta}

Since

\csc \theta = \frac{1}{\sin \theta}

, we have:

\frac{dy}{dx} = \frac{b}{a} \csc \theta

Final Answer: The derivative is

\frac{dy}{dx} = \frac{b}{a} \csc \theta

.

Q10EXERCISE 5.6

$x=a( heta+ heta an heta), y=a( an heta- heta an heta)$

Solution

Given: The parametric equations for a curve are:

x = a(\cos\theta + \theta\sin\theta)

y = a(\sin\theta - \theta\cos\theta)

To Find: The derivative

\frac{dy}{dx}

.

Solution: We will use parametric differentiation. First, we find the derivatives of

x

and

y

with respect to the parameter

\theta

.

Step 1: Differentiate

x

with respect to

\theta

.

x = a(\cos\theta + \theta\sin\theta)

\frac{dx}{d\theta} = a \frac{d}{d\theta}(\cos\theta + \theta\sin\theta)

\frac{dx}{d\theta} = a \left[ \frac{d}{d\theta}(\cos\theta) + \frac{d}{d\theta}(\theta\sin\theta) \right]

Using the product rule for

\theta\sin\theta

, we get

\frac{d}{d\theta}(\theta\sin\theta) = 1 \cdot \sin\theta + \theta \cdot \cos\theta = \sin\theta + \theta\cos\theta

.

\frac{dx}{d\theta} = a [-\sin\theta + (\sin\theta + \theta\cos\theta)]

\frac{dx}{d\theta} = a [-\sin\theta + \sin\theta + \theta\cos\theta] = a\theta\cos\theta

Step 2: Differentiate

y

with respect to

\theta

.

y = a(\sin\theta - \theta\cos\theta)

\frac{dy}{d\theta} = a \frac{d}{d\theta}(\sin\theta - \theta\cos\theta)

\frac{dy}{d\theta} = a \left[ \frac{d}{d\theta}(\sin\theta) - \frac{d}{d\theta}(\theta\cos\theta) \right]

Using the product rule for

\theta\cos\theta

, we get

\frac{d}{d\theta}(\theta\cos\theta) = 1 \cdot \cos\theta + \theta \cdot (-\sin\theta) = \cos\theta - \theta\sin\theta

.

\frac{dy}{d\theta} = a [\cos\theta - (\cos\theta - \theta\sin\theta)]

\frac{dy}{d\theta} = a [\cos\theta - \cos\theta + \theta\sin\theta] = a\theta\sin\theta

Step 3: Find

\frac{dy}{dx}

using the formula

\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}

.

\frac{dy}{dx} = \frac{a\theta\sin\theta}{a\theta\cos\theta}

\frac{dy}{dx} = \frac{\sin\theta}{\cos\theta} = \tan\theta

Final Answer: The derivative is

\frac{dy}{dx} = \tan\theta

.

Q11EXERCISE 5.6

If $x=\sqrt{a^{\sin ^{-1} t}}, y=\sqrt{a^{\cos ^{-1} t}}$ , show that $\frac{d y}{d x}=-\frac{y}{x}$

Solution

Given:

x = \sqrt{a^{\sin^{-1} t}}

y = \sqrt{a^{\cos^{-1} t}}

To Prove:

\frac{dy}{dx} = -\frac{y}{x}

Proof: We are given the parametric equations:

x = \sqrt{a^{\sin^{-1} t}} \quad ...(1)

y = \sqrt{a^{\cos^{-1} t}} \quad ...(2)

Squaring both equations, we get:

x^2 = a^{\sin^{-1} t} \quad ...(3)

y^2 = a^{\cos^{-1} t} \quad ...(4)

Now, multiply equation (3) and (4):

x^2 y^2 = a^{\sin^{-1} t} \times a^{\cos^{-1} t}

Using the property of exponents

a^m \times a^n = a^{m+n}

, we have:

x^2 y^2 = a^{(\sin^{-1} t + \cos^{-1} t)}

We know the identity for inverse trigonometric functions:

\sin^{-1} t + \cos^{-1} t = \frac{\pi}{2}

. Substituting this into the equation:

x^2 y^2 = a^{\frac{\pi}{2}}

Since

a^{\frac{\pi}{2}}

is a constant, we can differentiate the equation

x^2 y^2 = \text{constant}

with respect to

x

to find

\frac{dy}{dx}

.

Differentiating both sides with respect to

x

:

\frac{d}{dx}(x^2 y^2) = \frac{d}{dx}(a^{\frac{\pi}{2}})

Using the product rule on the left side and noting that the derivative of a constant is 0 on the right side:

\frac{d}{dx}(x^2) \cdot y^2 + x^2 \cdot \frac{d}{dx}(y^2) = 0

2x \cdot y^2 + x^2 \cdot \left(2y \frac{dy}{dx}\right) = 0

2x^2 y \frac{dy}{dx} = -2xy^2

Now, we solve for

\frac{dy}{dx}

(assuming

x, y \neq 0

):

\frac{dy}{dx} = -\frac{2xy^2}{2x^2 y}

\frac{dy}{dx} = -\frac{y}{x}

Hence Proved.

Q1EXERCISE 5.7

Find the second order derivatives of the functions given in Exercises 1 to 10 . $x^{2}+3 x+2$

Solution

Given: The function

f(x) = x^2 + 3x + 2

.

To Find: The second order derivative of the function.

Solution: Let

y = f(x) = x^2 + 3x + 2

.

Step 1: Find the first order derivative,

\frac{dy}{dx}

.

\frac{dy}{dx} = \frac{d}{dx}(x^2 + 3x + 2)

Using the power rule and sum rule for differentiation:

\frac{dy}{dx} = \frac{d}{dx}(x^2) + \frac{d}{dx}(3x) + \frac{d}{dx}(2)

\frac{dy}{dx} = 2x^{2-1} + 3x^{1-1} + 0

\frac{dy}{dx} = 2x + 3

Step 2: Find the second order derivative,

\frac{d^2y}{dx^2}

. The second order derivative is the derivative of the first order derivative.

\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}(2x + 3)

\frac{d^2y}{dx^2} = \frac{d}{dx}(2x) + \frac{d}{dx}(3)

\frac{d^2y}{dx^2} = 2 + 0

\frac{d^2y}{dx^2} = 2

Final Answer: The second order derivative of the function

x^2 + 3x + 2

is 2.

Q2EXERCISE 5.7

$x^{20}$

Solution

Given: The function

f(x) = x^{20}

.

To Find: The second order derivative of the function.

Solution: Let

y = f(x) = x^{20}

.

Step 1: Find the first order derivative,

\frac{dy}{dx}

. Using the power rule for differentiation,

\frac{d}{dx}(x^n) = nx^{n-1}

.

\frac{dy}{dx} = \frac{d}{dx}(x^{20})

\frac{dy}{dx} = 20x^{20-1}

\frac{dy}{dx} = 20x^{19}

Step 2: Find the second order derivative,

\frac{d^2y}{dx^2}

. The second order derivative is the derivative of the first order derivative.

\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}(20x^{19})

Applying the power rule again:

\frac{d^2y}{dx^2} = 20 \cdot \frac{d}{dx}(x^{19})

\frac{d^2y}{dx^2} = 20 \cdot (19x^{19-1})

\frac{d^2y}{dx^2} = 20 \cdot 19x^{18}

\frac{d^2y}{dx^2} = 380x^{18}

Final Answer: The second order derivative of the function

x^{20}

is

380x^{18}

.

Q3EXERCISE 5.7

$x \cdot \cos x$

Solution

Given: The function

y = x \cdot \cos x

.

To Find: The second derivative of the function,

\frac{d^2y}{dx^2}

.

Solution: We have the function

y = x \cos x

.

First, we find the first derivative,

\frac{dy}{dx}

, by applying the product rule, which states that

(uv)' = u'v + uv'

. Let

u = x

and

v = \cos x

. Then

u' = 1

and

v' = -\sin x

.

\frac{dy}{dx} = \frac{d}{dx}(x \cos x)

= (\frac{d}{dx}x) \cos x + x (\frac{d}{dx}\cos x)

= (1)\cos x + x(-\sin x)

= \cos x - x \sin x

Now, we find the second derivative,

\frac{d^2y}{dx^2}

, by differentiating

\frac{dy}{dx}

with respect to

x

.

\frac{d^2y}{dx^2} = \frac{d}{dx}(\cos x - x \sin x)

= \frac{d}{dx}(\cos x) - \frac{d}{dx}(x \sin x)

For the term

\frac{d}{dx}(x \sin x)

, we apply the product rule again. Let

u=x

and

v=\sin x

. Then

u'=1

and

v'=\cos x

.

\frac{d^2y}{dx^2} = -\sin x - [(\frac{d}{dx}x)\sin x + x(\frac{d}{dx}\sin x)]

= -\sin x - [(1)\sin x + x(\cos x)]

= -\sin x - (\sin x + x \cos x)

= -\sin x - \sin x - x \cos x

= -2 \sin x - x \cos x

Final Answer: The second derivative of

x \cdot \cos x

is

-2 \sin x - x \cos x

.

Q4EXERCISE 5.7

$\log x$

Solution

Given: The function

y = \log x

.

To Find: The second derivative of the function,

\frac{d^2y}{dx^2}

.

Solution: We have the function

y = \log x

.

First, we find the first derivative,

\frac{dy}{dx}

.

\frac{dy}{dx} = \frac{d}{dx}(\log x) = \frac{1}{x}

Now, we find the second derivative,

\frac{d^2y}{dx^2}

, by differentiating

\frac{dy}{dx}

with respect to

x

. We can write

\frac{1}{x}

as

x^{-1}

.

\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{1}{x}) = \frac{d}{dx}(x^{-1})

Using the power rule

\frac{d}{dx}(x^n) = nx^{n-1}

:

= -1 \cdot x^{-1-1}

= -x^{-2}

= -\frac{1}{x^2}

Final Answer: The second derivative of

\log x

is

-\frac{1}{x^2}

.

Q5EXERCISE 5.7

$x^{3} \log x$

Solution

Given: The function

y = x^3 \log x

.

To Find: The second derivative of the function,

\frac{d^2y}{dx^2}

.

Solution: We have the function

y = x^3 \log x

.

First, we find the first derivative,

\frac{dy}{dx}

, by applying the product rule,

(uv)' = u'v + uv'

. Let

u = x^3

and

v = \log x

. Then

u' = 3x^2

and

v' = \frac{1}{x}

.

\frac{dy}{dx} = \frac{d}{dx}(x^3 \log x)

= (\frac{d}{dx}x^3) \log x + x^3 (\frac{d}{dx}\log x)

= (3x^2)\log x + x^3(\frac{1}{x})

= 3x^2 \log x + x^2

Now, we find the second derivative,

\frac{d^2y}{dx^2}

, by differentiating

\frac{dy}{dx}

with respect to

x

.

\frac{d^2y}{dx^2} = \frac{d}{dx}(3x^2 \log x + x^2)

= \frac{d}{dx}(3x^2 \log x) + \frac{d}{dx}(x^2)

For the term

\frac{d}{dx}(3x^2 \log x)

, we apply the product rule again. Let

u=3x^2

and

v=\log x

. Then

u'=6x

and

v'=\frac{1}{x}

.

\frac{d^2y}{dx^2} = [(\frac{d}{dx}3x^2)\log x + 3x^2(\frac{d}{dx}\log x)] + 2x

= [(6x)\log x + 3x^2(\frac{1}{x})] + 2x

= 6x \log x + 3x + 2x

= 6x \log x + 5x

We can also write this as

x(6 \log x + 5)

.

Final Answer: The second derivative of

x^3 \log x

is

6x \log x + 5x

.

Q6EXERCISE 5.7

$e^{x} \sin 5 x$

Solution

Given: The function

y = e^x \sin 5x

.

To Find: The second derivative of the function,

\frac{d^2y}{dx^2}

.

Solution: We have the function

y = e^x \sin 5x

.

First, we find the first derivative,

\frac{dy}{dx}

, by applying the product rule,

(uv)' = u'v + uv'

. Let

u = e^x

and

v = \sin 5x

. Then

u' = e^x

and

v' = (\cos 5x) \cdot 5 = 5\cos 5x

.

\frac{dy}{dx} = \frac{d}{dx}(e^x \sin 5x)

= (\frac{d}{dx}e^x) \sin 5x + e^x (\frac{d}{dx}\sin 5x)

= e^x \sin 5x + e^x (5\cos 5x)

= e^x (\sin 5x + 5\cos 5x)

Now, we find the second derivative,

\frac{d^2y}{dx^2}

, by differentiating

\frac{dy}{dx}

with respect to

x

.

\frac{d^2y}{dx^2} = \frac{d}{dx}[e^x (\sin 5x + 5\cos 5x)]

We apply the product rule again. Let

u = e^x

and

v = \sin 5x + 5\cos 5x

. Then

u' = e^x

and

v' = \frac{d}{dx}(\sin 5x + 5\cos 5x) = 5\cos 5x + 5(-\sin 5x \cdot 5) = 5\cos 5x - 25\sin 5x

.

\frac{d^2y}{dx^2} = (\frac{d}{dx}e^x)(\sin 5x + 5\cos 5x) + e^x (\frac{d}{dx}(\sin 5x + 5\cos 5x))

= e^x (\sin 5x + 5\cos 5x) + e^x (5\cos 5x - 25\sin 5x)

= e^x [(\sin 5x + 5\cos 5x) + (5\cos 5x - 25\sin 5x)]

= e^x (\sin 5x - 25\sin 5x + 5\cos 5x + 5\cos 5x)

= e^x (-24\sin 5x + 10\cos 5x)

= 2e^x(5\cos 5x - 12\sin 5x)

Final Answer: The second derivative of

e^x \sin 5x

is

e^x(10\cos 5x - 24\sin 5x)

.

Q7EXERCISE 5.7

$e^{6 x} \cos 3 x$

Solution

Given: The function

y = e^{6x} \cos 3x

.

To Find: The second derivative of the function,

\frac{d^2y}{dx^2}

.

Solution: We have the function

y = e^{6x} \cos 3x

.

First, we find the first derivative,

\frac{dy}{dx}

, by applying the product rule,

(uv)' = u'v + uv'

. Let

u = e^{6x}

and

v = \cos 3x

. Then

u' = e^{6x} \cdot 6 = 6e^{6x}

and

v' = -\sin 3x \cdot 3 = -3\sin 3x

.

\frac{dy}{dx} = \frac{d}{dx}(e^{6x} \cos 3x)

= (\frac{d}{dx}e^{6x}) \cos 3x + e^{6x} (\frac{d}{dx}\cos 3x)

= (6e^{6x}) \cos 3x + e^{6x} (-3\sin 3x)

= e^{6x} (6\cos 3x - 3\sin 3x)

= 3e^{6x} (2\cos 3x - \sin 3x)

Now, we find the second derivative,

\frac{d^2y}{dx^2}

, by differentiating

\frac{dy}{dx}

with respect to

x

.

\frac{d^2y}{dx^2} = \frac{d}{dx}[e^{6x} (6\cos 3x - 3\sin 3x)]

We apply the product rule again. Let

u = e^{6x}

and

v = 6\cos 3x - 3\sin 3x

. Then

u' = 6e^{6x}

and

v' = \frac{d}{dx}(6\cos 3x - 3\sin 3x) = 6(-3\sin 3x) - 3(3\cos 3x) = -18\sin 3x - 9\cos 3x

.

\frac{d^2y}{dx^2} = (\frac{d}{dx}e^{6x})(6\cos 3x - 3\sin 3x) + e^{6x}(\frac{d}{dx}(6\cos 3x - 3\sin 3x))

= (6e^{6x})(6\cos 3x - 3\sin 3x) + e^{6x}(-18\sin 3x - 9\cos 3x)

= e^{6x} [6(6\cos 3x - 3\sin 3x) + (-18\sin 3x - 9\cos 3x)]

= e^{6x} [36\cos 3x - 18\sin 3x - 18\sin 3x - 9\cos 3x]

= e^{6x} [(36-9)\cos 3x - (18+18)\sin 3x]

= e^{6x} (27\cos 3x - 36\sin 3x)

= 9e^{6x} (3\cos 3x - 4\sin 3x)

Final Answer: The second derivative of

e^{6x} \cos 3x

is

9e^{6x}(3\cos 3x - 4\sin 3x)

.

Q8EXERCISE 5.7

$\tan ^{-1} x$

Solution

Given: The function is

y = \tan^{-1} x

.

To Find: The second derivative,

\frac{d^2y}{dx^2}

.

Solution: We have the function

y = \tan^{-1} x

.

First, we differentiate

y

with respect to

x

to find the first derivative,

\frac{dy}{dx}

.

\frac{dy}{dx} = \frac{d}{dx}(\tan^{-1} x)

Using the standard derivative of the inverse tangent function, we get:

\frac{dy}{dx} = \frac{1}{1 + x^2} = (1 + x^2)^{-1}

Next, we differentiate

\frac{dy}{dx}

with respect to

x

to find the second derivative,

\frac{d^2y}{dx^2}

.

\frac{d^2y}{dx^2} = \frac{d}{dx} \left( (1 + x^2)^{-1} \right)

Using the chain rule, where the outer function is

u^{-1}

and the inner function is

u = 1 + x^2

:

\frac{d^2y}{dx^2} = -1 \cdot (1 + x^2)^{-1-1} \cdot \frac{d}{dx}(1 + x^2)

\frac{d^2y}{dx^2} = -(1 + x^2)^{-2} \cdot (2x)

\frac{d^2y}{dx^2} = \frac{-2x}{(1 + x^2)^2}

Final Answer: The second derivative of

\tan^{-1} x

is

\frac{-2x}{(1 + x^2)^2}

.

Q9EXERCISE 5.7

$\log (\log x)$

Solution

Given: The function is

y = \log(\log x)

.

To Find: The second derivative,

\frac{d^2y}{dx^2}

.

Solution: We have the function

y = \log(\log x)

.

First, we differentiate

y

with respect to

x

to find the first derivative,

\frac{dy}{dx}

. Using the chain rule, let

u = \log x

. Then

y = \log u

.

\frac{dy}{dx} = \frac{d}{du}(\log u) \cdot \frac{du}{dx}

\frac{dy}{dx} = \frac{1}{u} \cdot \frac{d}{dx}(\log x)

Substituting back

u = \log x

:

\frac{dy}{dx} = \frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x \log x}

Next, we differentiate

\frac{dy}{dx}

with respect to

x

to find the second derivative,

\frac{d^2y}{dx^2}

. We use the quotient rule for differentiation, where

u=1

and

v=x \log x

.

\frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{1}{x \log x} \right) = \frac{(x \log x)\frac{d}{dx}(1) - 1 \cdot \frac{d}{dx}(x \log x)}{(x \log x)^2}

We know

\frac{d}{dx}(1) = 0

. For

\frac{d}{dx}(x \log x)

, we use the product rule:

\frac{d}{dx}(x \log x) = x \cdot \frac{d}{dx}(\log x) + (\log x) \cdot \frac{d}{dx}(x) = x \cdot \frac{1}{x} + \log x \cdot 1 = 1 + \log x

Now, substitute this back into the expression for the second derivative:

\frac{d^2y}{dx^2} = \frac{(x \log x)(0) - 1 \cdot (1 + \log x)}{(x \log x)^2}

\frac{d^2y}{dx^2} = \frac{-(1 + \log x)}{(x \log x)^2}

Final Answer: The second derivative of

\log(\log x)

is

\frac{-(1 + \log x)}{(x \log x)^2}

.

Q10EXERCISE 5.7

$\sin (\log x)$

Solution

Given: The function is

y = \sin(\log x)

.

To Find: The second derivative,

\frac{d^2y}{dx^2}

.

Solution: We have the function

y = \sin(\log x)

.

First, we differentiate

y

with respect to

x

to find the first derivative,

\frac{dy}{dx}

. Using the chain rule, let

u = \log x

. Then

y = \sin u

.

\frac{dy}{dx} = \frac{d}{du}(\sin u) \cdot \frac{du}{dx}

\frac{dy}{dx} = \cos u \cdot \frac{d}{dx}(\log x)

Substituting back

u = \log x

:

\frac{dy}{dx} = \cos(\log x) \cdot \frac{1}{x} = \frac{\cos(\log x)}{x}

Next, we differentiate

\frac{dy}{dx}

with respect to

x

to find the second derivative,

\frac{d^2y}{dx^2}

. We use the quotient rule, where

u = \cos(\log x)

and

v = x

.

\frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{\cos(\log x)}{x} \right) = \frac{x \cdot \frac{d}{dx}(\cos(\log x)) - \cos(\log x) \cdot \frac{d}{dx}(x)}{x^2}

First, find

\frac{d}{dx}(\cos(\log x))

using the chain rule:

\frac{d}{dx}(\cos(\log x)) = -\sin(\log x) \cdot \frac{d}{dx}(\log x) = -\sin(\log x) \cdot \frac{1}{x}

Now, substitute this into the expression for the second derivative:

\frac{d^2y}{dx^2} = \frac{x \left( -\frac{\sin(\log x)}{x} \right) - \cos(\log x) \cdot 1}{x^2}

\frac{d^2y}{dx^2} = \frac{-\sin(\log x) - \cos(\log x)}{x^2}

\frac{d^2y}{dx^2} = -\frac{\sin(\log x) + \cos(\log x)}{x^2}

Final Answer: The second derivative of

\sin(\log x)

is

-\frac{\sin(\log x) + \cos(\log x)}{x^2}

.

Q11EXERCISE 5.7

If $y=5 \cos x-3 \sin x$ , prove that $\frac{d^{2} y}{d x^{2}}+y=0$

Solution

Given:

y = 5 \cos x - 3 \sin x

.

To Prove:

\frac{d^2y}{dx^2} + y = 0

.

Proof: We are given the equation:

y = 5 \cos x - 3 \sin x \quad \dots(1)

First, we find the first derivative of

y

with respect to

x

:

\frac{dy}{dx} = \frac{d}{dx}(5 \cos x - 3 \sin x)

\frac{dy}{dx} = 5(-\sin x) - 3(\cos x)

\frac{dy}{dx} = -5 \sin x - 3 \cos x \quad \dots(2)

Next, we find the second derivative by differentiating

\frac{dy}{dx}

with respect to

x

:

\frac{d^2y}{dx^2} = \frac{d}{dx}(-5 \sin x - 3 \cos x)

\frac{d^2y}{dx^2} = -5(\cos x) - 3(-\sin x)

\frac{d^2y}{dx^2} = -5 \cos x + 3 \sin x

We can factor out

-1

from the expression:

\frac{d^2y}{dx^2} = -(5 \cos x - 3 \sin x)

From equation (1), we know that

y = 5 \cos x - 3 \sin x

. Substituting this into the equation for the second derivative:

\frac{d^2y}{dx^2} = -y

Now, we rearrange the equation to match the expression we need to prove:

\frac{d^2y}{dx^2} + y = 0

Hence Proved.

Q12EXERCISE 5.7

If $y=\cos ^{-1} x$ , Find $\frac{d^{2} y}{d x^{2}}$ in terms of $y$ alone.

Solution

Given:

y = \cos^{-1} x

.

To Find:

\frac{d^2y}{dx^2}

in terms of

y

alone.

Solution: We are given the function

y = \cos^{-1} x

. This implies

x = \cos y

.

First, we differentiate

y

with respect to

x

to find the first derivative,

\frac{dy}{dx}

.

\frac{dy}{dx} = \frac{d}{dx}(\cos^{-1} x)

Using the standard derivative of the inverse cosine function:

\frac{dy}{dx} = -\frac{1}{\sqrt{1-x^2}}

To express this in terms of

y

, we substitute

x = \cos y

into the expression. The principal value range for

y = \cos^{-1} x

is

[0, \pi]

. In this interval,

\sin y \ge 0

.

\sqrt{1-x^2} = \sqrt{1 - \cos^2 y} = \sqrt{\sin^2 y} = \sin y

So, the first derivative in terms of

y

is:

\frac{dy}{dx} = -\frac{1}{\sin y} = -\csc y

Next, we differentiate

\frac{dy}{dx}

with respect to

x

to find the second derivative,

\frac{d^2y}{dx^2}

.

\frac{d^2y}{dx^2} = \frac{d}{dx} \left( -\csc y \right)

We use the chain rule since we are differentiating a function of

y

with respect to

x

:

\frac{d^2y}{dx^2} = \frac{d}{dy}(-\csc y) \cdot \frac{dy}{dx}

The derivative of

-\csc y

with respect to

y

is

-(-\csc y \cot y) = \csc y \cot y

. So,

\frac{d^2y}{dx^2} = (\csc y \cot y) \cdot \frac{dy}{dx}

We already found that

\frac{dy}{dx} = -\csc y

. Substituting this into the equation:

\frac{d^2y}{dx^2} = (\csc y \cot y) \cdot (-\csc y)

\frac{d^2y}{dx^2} = -\csc^2 y \cot y

Final Answer: The second derivative

\frac{d^2y}{dx^2}

in terms of

y

alone is

-\csc^2 y \cot y

.

Q13EXERCISE 5.7

If $y=3 \cos (\log x)+4 \sin (\log x)$ , show that $x^{2} y_{2}+x y_{1}+y=0$

Solution

Given:

y = 3 \cos(\log x) + 4 \sin(\log x)

To Show:

x^2 y_2 + x y_1 + y = 0

Proof: We have the function:

y = 3 \cos(\log x) + 4 \sin(\log x)

Differentiating with respect to

x

, we get:

y_1 = \frac{dy}{dx} = 3 \cdot \left(-\sin(\log x) \cdot \frac{1}{x}\right) + 4 \cdot \left(\cos(\log x) \cdot \frac{1}{x}\right)

y_1 = \frac{-3 \sin(\log x) + 4 \cos(\log x)}{x}

Multiplying both sides by

x

:

x y_1 = -3 \sin(\log x) + 4 \cos(\log x)

Differentiating again with respect to

x

using the product rule on the left side:

\frac{d}{dx}(x y_1) = \frac{d}{dx}(-3 \sin(\log x) + 4 \cos(\log x))

1 \cdot y_1 + x \cdot y_2 = -3 \cdot \cos(\log x) \cdot \frac{1}{x} + 4 \cdot \left(-\sin(\log x) \cdot \frac{1}{x}\right)

x y_2 + y_1 = \frac{-3 \cos(\log x) - 4 \sin(\log x)}{x}

Multiplying both sides by

x

:

x(x y_2 + y_1) = -3 \cos(\log x) - 4 \sin(\log x)

x^2 y_2 + x y_1 = -(3 \cos(\log x) + 4 \sin(\log x))

From the given equation, we know that

y = 3 \cos(\log x) + 4 \sin(\log x)

. So, we can substitute

y

into the equation:

x^2 y_2 + x y_1 = -y

Rearranging the terms, we get:

x^2 y_2 + x y_1 + y = 0

Hence Proved.

Q14EXERCISE 5.7

If $y=\mathrm{A} e^{m x}+\mathrm{B} e^{n x}$ , show that $\frac{d^{2} y}{d x^{2}}-(m+n) \frac{d y}{d x}+m n y=0$

Solution

Given:

y = A e^{mx} + B e^{nx}

To Show:

\frac{d^2 y}{d x^2} - (m+n) \frac{dy}{dx} + mny = 0

Proof: We have the function:

y = A e^{mx} + B e^{nx} \quad ... (1)

Differentiating with respect to

x

, we get the first derivative:

\frac{dy}{dx} = A \cdot m e^{mx} + B \cdot n e^{nx} \quad ... (2)

Differentiating again with respect to

x

, we get the second derivative:

\frac{d^2 y}{d x^2} = A \cdot m^2 e^{mx} + B \cdot n^2 e^{nx} \quad ... (3)

Now, we substitute the expressions for

y

,

\frac{dy}{dx}

, and

\frac{d^2 y}{d x^2}

into the Left Hand Side (LHS) of the equation we need to prove: LHS =

\frac{d^2 y}{d x^2} - (m+n) \frac{dy}{dx} + mny

LHS =

(A m^2 e^{mx} + B n^2 e^{nx}) - (m+n)(A m e^{mx} + B n e^{nx}) + mn(A e^{mx} + B e^{nx})

Expanding the terms: LHS =

A m^2 e^{mx} + B n^2 e^{nx} - (m(A m e^{mx} + B n e^{nx}) + n(A m e^{mx} + B n e^{nx})) + A mn e^{mx} + B mn e^{nx}

LHS =

A m^2 e^{mx} + B n^2 e^{nx} - (A m^2 e^{mx} + B mn e^{mx} + A mn e^{mx} + B n^2 e^{nx}) + A mn e^{mx} + B mn e^{nx}

LHS =

A m^2 e^{mx} + B n^2 e^{nx} - A m^2 e^{mx} - B mn e^{mx} - A mn e^{mx} - B n^2 e^{nx} + A mn e^{mx} + B mn e^{nx}

Combining like terms: LHS =

(A m^2 e^{mx} - A m^2 e^{mx}) + (B n^2 e^{nx} - B n^2 e^{nx}) + (-B mn e^{mx} + B mn e^{mx}) + (-A mn e^{mx} + A mn e^{mx})

LHS =

0 + 0 + 0 + 0 = 0

Since LHS = 0 = RHS.

Hence Proved.

Q15EXERCISE 5.7

If $y=500 e^{7 x}+600 e^{-7 x}$ , show that $\frac{d^{2} y}{d x^{2}}=49 y$

Solution

Given:

y = 500 e^{7x} + 600 e^{-7x}

To Show:

\frac{d^2 y}{d x^2} = 49y

Proof: We have the function:

y = 500 e^{7x} + 600 e^{-7x}

Differentiating with respect to

x

, we get the first derivative:

\frac{dy}{dx} = 500 \cdot \frac{d}{dx}(e^{7x}) + 600 \cdot \frac{d}{dx}(e^{-7x})

\frac{dy}{dx} = 500 \cdot (7e^{7x}) + 600 \cdot (-7e^{-7x})

\frac{dy}{dx} = 3500 e^{7x} - 4200 e^{-7x}

Differentiating again with respect to

x

, we get the second derivative:

\frac{d^2 y}{d x^2} = 3500 \cdot \frac{d}{dx}(e^{7x}) - 4200 \cdot \frac{d}{dx}(e^{-7x})

\frac{d^2 y}{d x^2} = 3500 \cdot (7e^{7x}) - 4200 \cdot (-7e^{-7x})

\frac{d^2 y}{d x^2} = 24500 e^{7x} + 29400 e^{-7x}

Now, let's factor out 49 from the expression for the second derivative:

\frac{d^2 y}{d x^2} = 49 (\frac{24500}{49} e^{7x} + \frac{29400}{49} e^{-7x})

\frac{d^2 y}{d x^2} = 49 (500 e^{7x} + 600 e^{-7x})

From the given equation, we know that

y = 500 e^{7x} + 600 e^{-7x}

. Substituting this into our result:

\frac{d^2 y}{d x^2} = 49y

Hence Proved.

Q16EXERCISE 5.7

If $e^{y}(x+1)=1$ , show that $\frac{d^{2} y}{d x^{2}}=\left(\frac{d y}{d x}\right)^{2}$

Solution

Given:

e^y(x+1) = 1

To Show:

\frac{d^2 y}{d x^2} = \left(\frac{dy}{dx}\right)^2

Proof: We are given the relation:

e^y(x+1) = 1

We can write

e^y

as:

e^y = \frac{1}{x+1}

Method 1: Using implicit differentiation Differentiating the given relation

e^y(x+1) = 1

with respect to

x

using the product rule:

\frac{d}{dx}(e^y(x+1)) = \frac{d}{dx}(1)

e^y \frac{dy}{dx} (x+1) + e^y(1) = 0

Since

e^y \neq 0

, we can divide by

e^y

:

\frac{dy}{dx}(x+1) + 1 = 0

\frac{dy}{dx} = -\frac{1}{x+1}

Now, we differentiate this again with respect to

x

to find the second derivative:

\frac{d^2 y}{d x^2} = \frac{d}{dx} \left(-\frac{1}{x+1}\right) = -\frac{d}{dx}((x+1)^{-1})

\frac{d^2 y}{d x^2} = -(-1)(x+1)^{-2} \cdot (1)

\frac{d^2 y}{d x^2} = \frac{1}{(x+1)^2}

Now let's compute

\left(\frac{dy}{dx}\right)^2

:

\left(\frac{dy}{dx}\right)^2 = \left(-\frac{1}{x+1}\right)^2 = \frac{1}{(x+1)^2}

Comparing the two results, we see that:

\frac{d^2 y}{d x^2} = \left(\frac{dy}{dx}\right)^2

Method 2: Expressing y explicitly From

e^y = \frac{1}{x+1}

, we can take the natural logarithm of both sides:

y = \ln\left(\frac{1}{x+1}\right) = \ln(1) - \ln(x+1) = -\ln(x+1)

Now, we differentiate with respect to

x

:

\frac{dy}{dx} = -\frac{1}{x+1}

And differentiate again:

\frac{d^2 y}{d x^2} = -(-1)(x+1)^{-2} = \frac{1}{(x+1)^2}

Squaring the first derivative:

\left(\frac{dy}{dx}\right)^2 = \left(-\frac{1}{x+1}\right)^2 = \frac{1}{(x+1)^2}

Thus,

\frac{d^2 y}{d x^2} = \left(\frac{dy}{dx}\right)^2

.

Hence Proved.

Q17EXERCISE 5.7

If $y=\left(\tan ^{-1} x\right)^{2}$ , show that $\left(x^{2}+1\right)^{2} y_{2}+2 x\left(x^{2}+1\right) y_{1}=2$

Solution

Given:

y = (\tan^{-1} x)^2

To Show:

(x^2+1)^2 y_2 + 2x(x^2+1) y_1 = 2

Proof: We have the function:

y = (\tan^{-1} x)^2

Differentiating with respect to

x

, we get the first derivative,

y_1

:

y_1 = \frac{dy}{dx} = 2(\tan^{-1} x) \cdot \frac{d}{dx}(\tan^{-1} x)

y_1 = 2(\tan^{-1} x) \cdot \frac{1}{1+x^2}

To simplify, we multiply both sides by

(1+x^2)

:

(1+x^2)y_1 = 2 \tan^{-1} x

Now, we differentiate this equation again with respect to

x

. We use the product rule on the left side:

\frac{d}{dx}((1+x^2)y_1) = \frac{d}{dx}(2 \tan^{-1} x)

(\frac{d}{dx}(1+x^2)) \cdot y_1 + (1+x^2) \cdot (\frac{d}{dx}(y_1)) = 2 \cdot \frac{1}{1+x^2}

(2x)y_1 + (1+x^2)y_2 = \frac{2}{1+x^2}

Here,

y_2

represents the second derivative

\frac{d^2 y}{d x^2}

.

To eliminate the fraction, we multiply the entire equation by

(1+x^2)

:

(1+x^2) \left( 2xy_1 + (1+x^2)y_2 \right) = (1+x^2) \left( \frac{2}{1+x^2} \right)

2x(1+x^2)y_1 + (1+x^2)^2 y_2 = 2

Rearranging the terms to match the required format:

(x^2+1)^2 y_2 + 2x(x^2+1) y_1 = 2

Hence Proved.

Q1Miscellaneous Exercise on Chapter 5

Differentiate w.r.t. $x$ the function in Exercises 1 to 11. $\left(3 x^{2}-9 x+5\right)^{9}$

Solution

Given: The function

y = (3x^2 - 9x + 5)^9

.

To Find: The derivative of the function with respect to

x

.

Solution: We use the chain rule for differentiation, which states that if

y = f(g(x))

, then

\frac{dy}{dx} = f'(g(x)) \cdot g'(x)

.

Let

u = 3x^2 - 9x + 5

. Then

y = u^9

.

First, we differentiate

y

with respect to

u

:

\frac{dy}{du} = \frac{d}{du}(u^9) = 9u^8

Next, we differentiate

u

with respect to

x

:

\frac{du}{dx} = \frac{d}{dx}(3x^2 - 9x + 5) = 6x - 9

According to the chain rule,

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

. Substituting the expressions for

\frac{dy}{du}

and

\frac{du}{dx}

:

\frac{dy}{dx} = (9u^8) \cdot (6x - 9)

Now, substitute back

u = 3x^2 - 9x + 5

:

\frac{dy}{dx} = 9(3x^2 - 9x + 5)^8 (6x - 9)

We can factor out a 3 from the term

(6x - 9)

:

\frac{dy}{dx} = 9(3x^2 - 9x + 5)^8 \cdot 3(2x - 3)

\frac{dy}{dx} = 27(2x - 3)(3x^2 - 9x + 5)^8

Final Answer: The derivative of

(3x^2 - 9x + 5)^9

with respect to

x

is

27(2x - 3)(3x^2 - 9x + 5)^8

.

Q2Miscellaneous Exercise on Chapter 5

Differentiate w.r.t. $x$ the function in Exercises 1 to 11. 2. $\sin ^{3} x+\cos ^{6} x$

Solution

Given: The function

y = \sin^3 x + \cos^6 x

.

To Find: The derivative of the function with respect to

x

.

Solution: We use the sum rule for differentiation, which states that

\frac{d}{dx}(f(x) + g(x)) = \frac{d}{dx}f(x) + \frac{d}{dx}g(x)

.

\frac{dy}{dx} = \frac{d}{dx}(\sin^3 x) + \frac{d}{dx}(\cos^6 x)

For each term, we apply the chain rule.

For the first term,

\frac{d}{dx}(\sin^3 x)

: Let

u = \sin x

. Then we are differentiating

u^3

.

\frac{d}{dx}(\sin^3 x) = \frac{d}{dx}((\sin x)^3) = 3(\sin x)^2 \cdot \frac{d}{dx}(\sin x) = 3\sin^2 x \cdot \cos x

For the second term,

\frac{d}{dx}(\cos^6 x)

: Let

v = \cos x

. Then we are differentiating

v^6

.

\frac{d}{dx}(\cos^6 x) = \frac{d}{dx}((\cos x)^6) = 6(\cos x)^5 \cdot \frac{d}{dx}(\cos x) = 6\cos^5 x \cdot (-\sin x) = -6\cos^5 x \sin x

Now, we add the derivatives of the two terms:

\frac{dy}{dx} = 3\sin^2 x \cos x - 6\cos^5 x \sin x

We can factor out the common terms

3\sin x \cos x

:

\frac{dy}{dx} = 3\sin x \cos x (\sin x - 2\cos^4 x)

Final Answer: The derivative of

\sin^3 x + \cos^6 x

with respect to

x

is

3\sin^2 x \cos x - 6\cos^5 x \sin x

, which can also be written as

3\sin x \cos x (\sin x - 2\cos^4 x)

.

Q3Miscellaneous Exercise on Chapter 5

Differentiate w.r.t. $x$ the function in Exercises 1 to 11. 3. $(5 x)^{3 \cos 2 x}$

Solution

Given: The function

y = (5x)^{3\cos 2x}

.

To Find: The derivative of the function with respect to

x

.

Solution: This function is of the form

f(x)^{g(x)}

, so we use logarithmic differentiation.

Take the natural logarithm of both sides:

\ln y = \ln((5x)^{3\cos 2x})

Using the logarithm property

\ln(a^b) = b \ln a

:

\ln y = (3\cos 2x) \ln(5x)

Differentiate both sides with respect to

x

. For the left side, we use the chain rule:

\frac{d}{dx}(\ln y) = \frac{1}{y} \frac{dy}{dx}

For the right side, we use the product rule,

(uv)' = u'v + uv'

. Let

u = 3\cos 2x

and

v = \ln(5x)

.

First, find the derivatives of

u

and

v

:

\frac{du}{dx} = \frac{d}{dx}(3\cos 2x) = 3(-\sin 2x) \cdot 2 = -6\sin 2x

\frac{dv}{dx} = \frac{d}{dx}(\ln(5x)) = \frac{1}{5x} \cdot 5 = \frac{1}{x}

Applying the product rule:

\frac{d}{dx}((3\cos 2x) \ln(5x)) = (-6\sin 2x) \ln(5x) + (3\cos 2x) \left(\frac{1}{x}\right)

= \frac{3\cos 2x}{x} - 6\sin 2x \ln(5x)

Now, equate the derivatives of both sides:

\frac{1}{y} \frac{dy}{dx} = \frac{3\cos 2x}{x} - 6\sin 2x \ln(5x)

Solve for

\frac{dy}{dx}

by multiplying by

y

:

\frac{dy}{dx} = y \left( \frac{3\cos 2x}{x} - 6\sin 2x \ln(5x) \right)

Substitute back

y = (5x)^{3\cos 2x}

:

\frac{dy}{dx} = (5x)^{3\cos 2x} \left( \frac{3\cos 2x}{x} - 6\sin 2x \ln(5x) \right)

Final Answer: The derivative of

(5x)^{3\cos 2x}

with respect to

x

is

(5x)^{3\cos 2x} \left( \frac{3\cos 2x}{x} - 6\sin 2x \ln(5x) \right)

.

Q4Miscellaneous Exercise on Chapter 5

Differentiate w.r.t. $x$ the function in Exercises 1 to 11. 4. $\sin ^{-1}(x \sqrt{x}), 0 \leq x \leq 1$

Solution

Given: The function

y = \sin^{-1}(x\sqrt{x})

, for

0 \leq x \leq 1

.

To Find: The derivative of the function with respect to

x

.

Solution: First, simplify the expression inside the inverse sine function:

x\sqrt{x} = x^1 \cdot x^{1/2} = x^{1 + 1/2} = x^{3/2}

So, the function is

y = \sin^{-1}(x^{3/2})

.

We use the chain rule for differentiation. Let

u = x^{3/2}

. Then

y = \sin^{-1}(u)

. The derivative of

\sin^{-1}(u)

with respect to

u

is

\frac{1}{\sqrt{1-u^2}}

.

According to the chain rule,

\frac{dy}{dx} = \frac{d}{du}(\sin^{-1} u) \cdot \frac{du}{dx}

.

First, find

\frac{du}{dx}

:

\frac{du}{dx} = \frac{d}{dx}(x^{3/2}) = \frac{3}{2}x^{(3/2 - 1)} = \frac{3}{2}x^{1/2} = \frac{3\sqrt{x}}{2}

Now, apply the chain rule:

\frac{dy}{dx} = \frac{1}{\sqrt{1-u^2}} \cdot \frac{3\sqrt{x}}{2}

Substitute back

u = x^{3/2}

. Then

u^2 = (x^{3/2})^2 = x^3

.

\frac{dy}{dx} = \frac{1}{\sqrt{1-x^3}} \cdot \frac{3\sqrt{x}}{2}

Combine the terms:

\frac{dy}{dx} = \frac{3\sqrt{x}}{2\sqrt{1-x^3}}

The condition

0 \leq x \leq 1

ensures that the function and its derivative are well-defined (for

x=1

, the denominator of the derivative is 0, so the derivative is undefined at that endpoint).

Final Answer: The derivative of

\sin^{-1}(x\sqrt{x})

with respect to

x

is

\frac{3\sqrt{x}}{2\sqrt{1-x^3}}

.

Q5Miscellaneous Exercise on Chapter 5

Differentiate w.r.t. $x$ the function in Exercises 1 to 11. 5. $\frac{\cos ^{-1} \frac{x}{2}}{\sqrt{2 x+7}}, -2<x<2$

Solution

Given: The function

y = \frac{\cos^{-1} \frac{x}{2}}{\sqrt{2x+7}}

, for

-2 < x < 2

.

To Find: The derivative of the function with respect to

x

.

Solution: We use the quotient rule, which states

\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}

.

Let

u = \cos^{-1}\left(\frac{x}{2}\right)

and

v = \sqrt{2x+7}

.

First, we find

\frac{du}{dx}

using the chain rule. The derivative of

\cos^{-1}(t)

is

\frac{-1}{\sqrt{1-t^2}}

.

\frac{du}{dx} = \frac{d}{dx}\left(\cos^{-1}\left(\frac{x}{2}\right)\right) = \frac{-1}{\sqrt{1 - (x/2)^2}} \cdot \frac{d}{dx}\left(\frac{x}{2}\right)

\frac{du}{dx} = \frac{-1}{\sqrt{1 - x^2/4}} \cdot \frac{1}{2} = \frac{-1}{2\sqrt{\frac{4-x^2}{4}}} = \frac{-1}{2 \frac{\sqrt{4-x^2}}{2}} = \frac{-1}{\sqrt{4-x^2}}

Next, we find

\frac{dv}{dx}

using the chain rule:

v = (2x+7)^{1/2}

\frac{dv}{dx} = \frac{1}{2}(2x+7)^{-1/2} \cdot \frac{d}{dx}(2x+7) = \frac{1}{2\sqrt{2x+7}} \cdot 2 = \frac{1}{\sqrt{2x+7}}

Now, we apply the quotient rule:

\frac{dy}{dx} = \frac{(\sqrt{2x+7})\left(\frac{-1}{\sqrt{4-x^2}}\right) - \left(\cos^{-1}\frac{x}{2}\right)\left(\frac{1}{\sqrt{2x+7}}\right)}{(\sqrt{2x+7})^2}

\frac{dy}{dx} = \frac{\frac{-\sqrt{2x+7}}{\sqrt{4-x^2}} - \frac{\cos^{-1}(x/2)}{\sqrt{2x+7}}}{2x+7}

To simplify the numerator, we find a common denominator, which is

\sqrt{4-x^2}\sqrt{2x+7}

:

\text{Numerator} = \frac{-(\sqrt{2x+7})(\sqrt{2x+7}) - (\sqrt{4-x^2})(\cos^{-1}(x/2))}{\sqrt{4-x^2}\sqrt{2x+7}}

\text{Numerator} = \frac{-(2x+7) - \sqrt{4-x^2}\cos^{-1}(x/2)}{\sqrt{4-x^2}\sqrt{2x+7}}

Now, substitute this back into the expression for

\frac{dy}{dx}

:

\frac{dy}{dx} = \frac{-(2x+7) - \sqrt{4-x^2}\cos^{-1}(x/2)}{(\sqrt{4-x^2}\sqrt{2x+7})(2x+7)}

Combine the terms in the denominator:

\frac{dy}{dx} = \frac{-(2x+7) - \sqrt{4-x^2}\cos^{-1}(x/2)}{\sqrt{4-x^2}(2x+7)^{3/2}}

We can also split the fraction:

\frac{dy}{dx} = -\frac{2x+7}{\sqrt{4-x^2}(2x+7)^{3/2}} - \frac{\sqrt{4-x^2}\cos^{-1}(x/2)}{\sqrt{4-x^2}(2x+7)^{3/2}}

\frac{dy}{dx} = -\frac{1}{\sqrt{4-x^2}\sqrt{2x+7}} - \frac{\cos^{-1}(x/2)}{(2x+7)^{3/2}}

Final Answer: The derivative of

\frac{\cos^{-1} \frac{x}{2}}{\sqrt{2x+7}}

with respect to

x

is

-\frac{1}{\sqrt{4-x^2}\sqrt{2x+7}} - \frac{\cos^{-1}(x/2)}{(2x+7)^{3/2}}

.

Q6Miscellaneous Exercise on Chapter 5

$\cot ^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right], 0<x<\frac{\pi}{2}$

Solution

Given: The function

y = \cot ^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]

, for

0 0

,

\sin\frac{x}{2} > 0

, and also

\cos\frac{x}{2} > \sin\frac{x}{2}

. Therefore:

\cos\frac{x}{2} + \sin\frac{x}{2} > 0 \implies \left|\cos\frac{x}{2} + \sin\frac{x}{2}\right| = \cos\frac{x}{2} + \sin\frac{x}{2}

\cos\frac{x}{2} - \sin\frac{x}{2} > 0 \implies \left|\cos\frac{x}{2} - \sin\frac{x}{2}\right| = \cos\frac{x}{2} - \sin\frac{x}{2}

Now, we substitute these simplified terms back into the expression for

y

: The numerator is:

\sqrt{1+\sin x}+\sqrt{1-\sin x} = \left(\cos\frac{x}{2} + \sin\frac{x}{2}\right) + \left(\cos\frac{x}{2} - \sin\frac{x}{2}\right) = 2\cos\frac{x}{2}

The denominator is:

\sqrt{1+\sin x}-\sqrt{1-\sin x} = \left(\cos\frac{x}{2} + \sin\frac{x}{2}\right) - \left(\cos\frac{x}{2} - \sin\frac{x}{2}\right) = 2\sin\frac{x}{2}

The expression inside

\cot^{-1}

becomes:

\frac{2\cos\frac{x}{2}}{2\sin\frac{x}{2}} = \cot\frac{x}{2}

So,

y = \cot^{-1}\left(\cot\frac{x}{2}\right)

. Since

0 < \frac{x}{2} < \frac{\pi}{4}

, which is within the principal value range of

\cot^{-1}(t)

(which is

(0, \pi)

), we can write:

y = \frac{x}{2}

Now, we differentiate

y

with respect to

x

:

\frac{dy}{dx} = \frac{d}{dx}\left(\frac{x}{2}\right) = \frac{1}{2}

Final Answer: The derivative of the given function is

\frac{1}{2}

.

Q7Miscellaneous Exercise on Chapter 5

$(\log x)^{\log x}, x>1$

Solution

Given: The function

y = (\log x)^{\log x}

, for

x>1

.

To Find: The derivative

\frac{dy}{dx}

.

Solution: The function is of the form

f(x)^{g(x)}

, so we use logarithmic differentiation. Taking the natural logarithm on both sides of the equation

y = (\log x)^{\log x}

:

\log y = \log\left[(\log x)^{\log x}\right]

Using the logarithm property

\log(a^b) = b \log a

, we get:

\log y = (\log x) \cdot \log(\log x)

Now, we differentiate both sides with respect to

x

. Using the product rule on the right side:

\frac{d}{dx}(\log y) = \frac{d}{dx}[(\log x) \cdot \log(\log x)]

\frac{1}{y}\frac{dy}{dx} = \left[\frac{d}{dx}(\log x)\right] \cdot \log(\log x) + (\log x) \cdot \left[\frac{d}{dx}(\log(\log x))\right]

Applying the chain rule:

\frac{1}{y}\frac{dy}{dx} = \frac{1}{x} \cdot \log(\log x) + (\log x) \cdot \frac{1}{\log x} \cdot \frac{d}{dx}(\log x)

\frac{1}{y}\frac{dy}{dx} = \frac{\log(\log x)}{x} + (\log x) \cdot \frac{1}{\log x} \cdot \frac{1}{x}

\frac{1}{y}\frac{dy}{dx} = \frac{\log(\log x)}{x} + \frac{1}{x}

\frac{1}{y}\frac{dy}{dx} = \frac{1 + \log(\log x)}{x}

Now, we solve for

\frac{dy}{dx}

by multiplying both sides by

y

:

\frac{dy}{dx} = y \left[ \frac{1 + \log(\log x)}{x} \right]

Finally, substitute back

y = (\log x)^{\log x}

:

\frac{dy}{dx} = (\log x)^{\log x} \left[ \frac{1 + \log(\log x)}{x} \right]

Final Answer: The derivative of

(\log x)^{\log x}

with respect to

x

is

(\log x)^{\log x} \left[ \frac{1 + \log(\log x)}{x} \right]

.

Q8Miscellaneous Exercise on Chapter 5

$\cos (a \cos x+b \sin x)$ , for some constant $a$ and $b$ .

Solution

Given: The function

y = \cos(a \cos x + b \sin x)

, for some constants

a

and

b

.

To Find: The derivative

\frac{dy}{dx}

.

Solution: We will use the chain rule to differentiate the function. Let the inner function be

u = a \cos x + b \sin x

. Then the function becomes

y = \cos u

.

According to the chain rule,

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

.

First, we find the derivative of the outer function

y

with respect to

u

:

\frac{dy}{du} = \frac{d}{du}(\cos u) = -\sin u

Substituting

u

back, we get:

\frac{dy}{du} = -\sin(a \cos x + b \sin x)

Next, we find the derivative of the inner function

u

with respect to

x

:

\frac{du}{dx} = \frac{d}{dx}(a \cos x + b \sin x)

\frac{du}{dx} = a \frac{d}{dx}(\cos x) + b \frac{d}{dx}(\sin x)

\frac{du}{dx} = a(-\sin x) + b(\cos x) = b \cos x - a \sin x

Now, we multiply the two derivatives to find

\frac{dy}{dx}

:

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

\frac{dy}{dx} = [-\sin(a \cos x + b \sin x)] \cdot (b \cos x - a \sin x)

\frac{dy}{dx} = -(b \cos x - a \sin x) \sin(a \cos x + b \sin x)

\frac{dy}{dx} = (a \sin x - b \cos x) \sin(a \cos x + b \sin x)

Final Answer: The derivative of

\cos(a \cos x + b \sin x)

with respect to

x

is

(a \sin x - b \cos x) \sin(a \cos x + b \sin x)

.

Q9Miscellaneous Exercise on Chapter 5

$(\sin x-\cos x)^{(\sin x-\cos x)}, \frac{\pi}{4}<x<\frac{3 \pi}{4}$

Solution

Given: The function

y = (\sin x - \cos x)^{(\sin x - \cos x)}

, for

\frac{\pi}{4} < x < \frac{3\pi}{4}

.

To Find: The derivative

\frac{dy}{dx}

.

Solution: The given domain

\frac{\pi}{4} < x < \frac{3\pi}{4}

ensures that

\sin x > \cos x

, so the base

(\sin x - \cos x)

is positive. This allows us to use logarithmic differentiation.

Let

y = (\sin x - \cos x)^{(\sin x - \cos x)}

. Taking the natural logarithm on both sides:

\log y = \log\left[(\sin x - \cos x)^{(\sin x - \cos x)}\right]

Using the logarithm property

\log(a^b) = b \log a

, we get:

\log y = (\sin x - \cos x) \log(\sin x - \cos x)

Now, we differentiate both sides with respect to

x

. We use the product rule on the right side.

\frac{d}{dx}(\log y) = \frac{d}{dx}[(\sin x - \cos x) \log(\sin x - \cos x)]

\frac{1}{y}\frac{dy}{dx} = \left[\frac{d}{dx}(\sin x - \cos x)\right] \log(\sin x - \cos x) + (\sin x - \cos x) \left[\frac{d}{dx}\log(\sin x - \cos x)\right]

First, let's find the derivative of

(\sin x - \cos x)

:

\frac{d}{dx}(\sin x - \cos x) = \cos x - (-\sin x) = \cos x + \sin x

Now, substitute this into the differentiation equation:

\frac{1}{y}\frac{dy}{dx} = (\cos x + \sin x) \log(\sin x - \cos x) + (\sin x - \cos x) \cdot \frac{1}{\sin x - \cos x} \cdot \frac{d}{dx}(\sin x - \cos x)

\frac{1}{y}\frac{dy}{dx} = (\cos x + \sin x) \log(\sin x - \cos x) + (\sin x - \cos x) \cdot \frac{1}{\sin x - \cos x} \cdot (\cos x + \sin x)

\frac{1}{y}\frac{dy}{dx} = (\cos x + \sin x) \log(\sin x - \cos x) + (\cos x + \sin x)

Factor out the common term

(\cos x + \sin x)

:

\frac{1}{y}\frac{dy}{dx} = (\cos x + \sin x) [1 + \log(\sin x - \cos x)]

To find

\frac{dy}{dx}

, multiply by

y

:

\frac{dy}{dx} = y (\cos x + \sin x) [1 + \log(\sin x - \cos x)]

Finally, substitute back

y = (\sin x - \cos x)^{(\sin x - \cos x)}

:

\frac{dy}{dx} = (\sin x - \cos x)^{(\sin x - \cos x)} (\cos x + \sin x) [1 + \log(\sin x - \cos x)]

Final Answer: The derivative of

(\sin x - \cos x)^{(\sin x - \cos x)}

is

(\sin x - \cos x)^{(\sin x - \cos x)} (\cos x + \sin x) [1 + \log(\sin x - \cos x)]

.

Q10Miscellaneous Exercise on Chapter 5

$x^{x}+x^{a}+a^{x}+a^{a}$ , for some fixed $a>0$ and $x>0$

Solution

Given: The function

y = x^x + x^a + a^x + a^a

, for some fixed

a>0

and

x>0

.

To Find: The derivative

\frac{dy}{dx}

.

Solution: We can differentiate each term of the sum separately.

\frac{dy}{dx} = \frac{d}{dx}(x^x) + \frac{d}{dx}(x^a) + \frac{d}{dx}(a^x) + \frac{d}{dx}(a^a)

Let's find the derivative of each term:

1. Derivative of $x^x$ : Let

u = x^x

. We use logarithmic differentiation.

\log u = \log(x^x) = x \log x

. Differentiating with respect to

x

:

\frac{1}{u}\frac{du}{dx} = \frac{d}{dx}(x \log x) = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1

\frac{du}{dx} = u(1 + \log x) = x^x(1 + \log x)

So,

\frac{d}{dx}(x^x) = x^x(1 + \log x)

.

2. Derivative of $x^a$ : This is a power function. We use the power rule

\frac{d}{dx}(x^n) = nx^{n-1}

.

\frac{d}{dx}(x^a) = a x^{a-1}

3. Derivative of $a^x$ : This is an exponential function. We use the rule

\frac{d}{dx}(c^x) = c^x \log c

.

\frac{d}{dx}(a^x) = a^x \log a

4. Derivative of $a^a$ : Since

a

is a fixed constant,

a^a

is also a constant. The derivative of a constant is 0.

\frac{d}{dx}(a^a) = 0

Now, we combine the derivatives of all the terms:

\frac{dy}{dx} = x^x(1 + \log x) + a x^{a-1} + a^x \log a + 0

Final Answer: The derivative of

x^x + x^a + a^x + a^a

with respect to

x

is

x^x(1 + \log x) + ax^{a-1} + a^x \log a

.

Q11Miscellaneous Exercise on Chapter 5

$x^{x^{2}-3}+(x-3)^{x^{2}}$ , for $x>3$

Solution

Given: The function

f(x) = x^{x^{2}-3}+(x-3)^{x^{2}}

, for

x>3

.

To Find: The derivative of the function with respect to

x

.

Solution: Let the given function be

y = x^{x^{2}-3}+(x-3)^{x^{2}}

. This function is a sum of two functions of the form

f(x)^{g(x)}

. We will differentiate each term separately using logarithmic differentiation.

Let

u = x^{x^{2}-3}

and

v = (x-3)^{x^{2}}

. Then

y = u + v

, and

\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}

.

First, consider

u = x^{x^{2}-3}

. Taking the natural logarithm on both sides:

\ln u = \ln(x^{x^{2}-3})

\ln u = (x^2 - 3) \ln x

Differentiating with respect to

x

using the product rule:

\frac{1}{u} \frac{du}{dx} = \frac{d}{dx}(x^2 - 3) \cdot \ln x + (x^2 - 3) \cdot \frac{d}{dx}(\ln x)

\frac{1}{u} \frac{du}{dx} = (2x) \ln x + (x^2 - 3) \frac{1}{x}

\frac{1}{u} \frac{du}{dx} = 2x \ln x + x - \frac{3}{x}

\frac{du}{dx} = u \left( 2x \ln x + x - \frac{3}{x} \right)

Substituting back

u = x^{x^{2}-3}

:

\frac{du}{dx} = x^{x^{2}-3} \left( 2x \ln x + x - \frac{3}{x} \right)

Next, consider

v = (x-3)^{x^{2}}

. Taking the natural logarithm on both sides:

\ln v = \ln((x-3)^{x^{2}})

\ln v = x^2 \ln(x-3)

Differentiating with respect to

x

using the product rule:

\frac{1}{v} \frac{dv}{dx} = \frac{d}{dx}(x^2) \cdot \ln(x-3) + x^2 \cdot \frac{d}{dx}(\ln(x-3))

\frac{1}{v} \frac{dv}{dx} = (2x) \ln(x-3) + x^2 \left( \frac{1}{x-3} \right)

\frac{dv}{dx} = v \left( 2x \ln(x-3) + \frac{x^2}{x-3} \right)

Substituting back

v = (x-3)^{x^{2}}

:

\frac{dv}{dx} = (x-3)^{x^{2}} \left( 2x \ln(x-3) + \frac{x^2}{x-3} \right)

Finally, we add the derivatives

\frac{du}{dx}

and

\frac{dv}{dx}

to find

\frac{dy}{dx}

.

\frac{dy}{dx} = x^{x^{2}-3} \left( 2x \ln x + x - \frac{3}{x} \right) + (x-3)^{x^{2}} \left( 2x \ln(x-3) + \frac{x^2}{x-3} \right)

Final Answer: The derivative is

x^{x^{2}-3} \left( 2x \ln x + x - \frac{3}{x} \right) + (x-3)^{x^{2}} \left( 2x \ln(x-3) + \frac{x^2}{x-3} \right)

.

Q12Miscellaneous Exercise on Chapter 5

Find $\frac{d y}{d x}$ , if $y=12(1-\cos t), x=10(t-\sin t),-\frac{\pi}{2}<t<\frac{\pi}{2}$

Solution

Given: The parametric equations for a curve are:

y = 12(1 - \cos t)

x = 10(t - \sin t)

for

-\frac{\pi}{2} < t < \frac{\pi}{2}

.

To Find: The derivative

\frac{dy}{dx}

.

Formula: For parametric equations

x = f(t)

and

y = g(t)

, the derivative is given by:

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}

Solution: First, we differentiate

y

with respect to

t

:

y = 12(1 - \cos t)

\frac{dy}{dt} = 12 \frac{d}{dt}(1 - \cos t)

\frac{dy}{dt} = 12(0 - (-\sin t)) = 12 \sin t

Next, we differentiate

x

with respect to

t

:

x = 10(t - \sin t)

\frac{dx}{dt} = 10 \frac{d}{dt}(t - \sin t)

\frac{dx}{dt} = 10(1 - \cos t)

Now, we find

\frac{dy}{dx}

using the formula for parametric differentiation:

\frac{dy}{dx} = \frac{12 \sin t}{10(1 - \cos t)} = \frac{6 \sin t}{5(1 - \cos t)}

To simplify the expression, we use the half-angle trigonometric identities:

\sin t = 2 \sin(t/2) \cos(t/2)

1 - \cos t = 2 \sin^2(t/2)

Substituting these into the expression for

\frac{dy}{dx}

:

\frac{dy}{dx} = \frac{6 \cdot (2 \sin(t/2) \cos(t/2))}{5 \cdot (2 \sin^2(t/2))}

\frac{dy}{dx} = \frac{6 \cos(t/2)}{5 \sin(t/2)}

\frac{dy}{dx} = \frac{6}{5} \cot(t/2)

For the given range

-\frac{\pi}{2} < t < \frac{\pi}{2}

and

t \neq 0

, the denominator

10(1-\cos t)

is not zero.

Final Answer:

\frac{dy}{dx} = \frac{6}{5} \cot(t/2)

.

Q13Miscellaneous Exercise on Chapter 5

Find $\frac{d y}{d x}$ , if $y=\sin ^{-1} x+\sin ^{-1} \sqrt{1-x^{2}}, 0<x<1$

Solution

Given: The function

y = \sin^{-1} x + \sin^{-1} \sqrt{1-x^2}

, for

0 < x < 1

.

To Find: The derivative

\frac{dy}{dx}

.

Solution: We can simplify the given function using a trigonometric substitution. Let

x = \sin \theta

. Since

0 < x < 1

, we have

0 < \sin \theta < 1

, which implies

0 < \theta < \frac{\pi}{2}

.

Now substitute

x = \sin \theta

into the expression for

y

:

y = \sin^{-1}(\sin \theta) + \sin^{-1} \sqrt{1-(\sin \theta)^2}

y = \sin^{-1}(\sin \theta) + \sin^{-1} \sqrt{\cos^2 \theta}

Since

0 < \theta < \frac{\pi}{2}

,

\cos \theta

is positive. So,

\sqrt{\cos^2 \theta} = \cos \theta

.

y = \sin^{-1}(\sin \theta) + \sin^{-1}(\cos \theta)

For

0 < \theta < \frac{\pi}{2}

, we have

\sin^{-1}(\sin \theta) = \theta

. Also, we can write

\cos \theta = \sin(\frac{\pi}{2} - \theta)

. Since

0 < \theta < \frac{\pi}{2}

, we have

0 < \frac{\pi}{2} - \theta < \frac{\pi}{2}

. The range of

\sin^{-1}

is

[-\frac{\pi}{2}, \frac{\pi}{2}]

, so we can write:

\sin^{-1}(\cos \theta) = \sin^{-1}\left(\sin\left(\frac{\pi}{2} - \theta\right)\right) = \frac{\pi}{2} - \theta

Substituting these back into the expression for

y

:

y = \theta + \left(\frac{\pi}{2} - \theta\right)

y = \frac{\pi}{2}

The function

y

is a constant.

Now, we differentiate

y

with respect to

x

:

\frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{2}\right)

\frac{dy}{dx} = 0

Alternative Method (Direct Differentiation):

y = \sin^{-1} x + \sin^{-1} \sqrt{1-x^2}

\frac{dy}{dx} = \frac{d}{dx}(\sin^{-1} x) + \frac{d}{dx}(\sin^{-1} \sqrt{1-x^2})

\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} + \frac{1}{\sqrt{1-(\sqrt{1-x^2})^2}} \cdot \frac{d}{dx}(\sqrt{1-x^2})

\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} + \frac{1}{\sqrt{1-(1-x^2)}} \cdot \frac{-2x}{2\sqrt{1-x^2}}

\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} + \frac{1}{\sqrt{x^2}} \cdot \frac{-x}{\sqrt{1-x^2}}

Since

0 < x < 1

,

\sqrt{x^2} = x

.

\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} + \frac{1}{x} \cdot \frac{-x}{\sqrt{1-x^2}}

\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} - \frac{1}{\sqrt{1-x^2}} = 0

Final Answer:

\frac{dy}{dx} = 0

.

Q14Miscellaneous Exercise on Chapter 5

If $x \sqrt{1+y}+y \sqrt{1+x}=0$ , for , $-1<x<1$ , prove that $\frac{d y}{d x}=-\frac{1}{(1+x)^{2}}$

Solution

Given: The equation

x \sqrt{1+y} + y \sqrt{1+x} = 0

, for

-1 < x < 1

.

To Prove:

\frac{dy}{dx} = -\frac{1}{(1+x)^2}

.

Proof: We are given the equation:

x \sqrt{1+y} + y \sqrt{1+x} = 0

First, we will solve this equation for

y

in terms of

x

.

x \sqrt{1+y} = -y \sqrt{1+x}

Squaring both sides of the equation:

x^2 (1+y) = (-y)^2 (1+x)

x^2 + x^2 y = y^2 + y^2 x

Rearrange the terms to group

x

and

y

terms:

x^2 - y^2 = y^2 x - x^2 y

Factor both sides:

(x-y)(x+y) = -xy(x-y)

We have two possibilities:

$x - y = 0 \implies x = y$ . If we substitute $y=x$ in the original equation, we get $x\sqrt{1+x} + x\sqrt{1+x} = 0$ , which means $2x\sqrt{1+x}=0$ . This is true if $x=0$ or $x=-1$ . Since the domain is $-1<x<1$ , we have $x=y=0$ . This is a specific solution, not a general relation for the given interval. If $y=x$ , then $\frac{dy}{dx}=1$ , which is not what we want to prove. So we consider the case $x \neq y$ .
If $x \neq y$ , we can divide both sides by $(x-y)$ : $x+y = -xy$ Now, we solve for $y$ : $y + xy = -x$ $y(1+x) = -x$ $y = -\frac{x}{1+x}$

Now we differentiate this expression for

y

with respect to

x

using the quotient rule, which states

\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}

. Let

u=x

and

v=1+x

. Then

u'=1

and

v'=1

.

\frac{dy}{dx} = - \left[ \frac{\frac{d}{dx}(x) \cdot (1+x) - x \cdot \frac{d}{dx}(1+x)}{(1+x)^2} \right]

\frac{dy}{dx} = - \left[ \frac{1 \cdot (1+x) - x \cdot 1}{(1+x)^2} \right]

\frac{dy}{dx} = - \left[ \frac{1+x - x}{(1+x)^2} \right]

\frac{dy}{dx} = - \frac{1}{(1+x)^2}

This is the required result.

Hence Proved.

Q15Miscellaneous Exercise on Chapter 5

If $(x-a)^{2}+(y-b)^{2}=c^{2}$ , for some $c>0$ , prove that $\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{\frac{3}{2}}}{\frac{d^{2} y}{d x^{2}}}$ is a constant independent of $a$ and $b$ .

Solution

Given: The equation

(x-a)^2 + (y-b)^2 = c^2

, for some constant

c > 0

.

To Prove: The expression

\frac{\left[1+\left(\frac{dy}{dx}\right)^{2}\right]^{\frac{3}{2}}}{\frac{d^{2} y}{d x^{2}}}

is a constant independent of

a

and

b

.

Proof: The given equation represents a circle with center

(a, b)

and radius

c

.

(x-a)^2 + (y-b)^2 = c^2

We differentiate this equation implicitly with respect to

x

.

\frac{d}{dx}[(x-a)^2] + \frac{d}{dx}[(y-b)^2] = \frac{d}{dx}[c^2]

2(x-a) \cdot 1 + 2(y-b) \cdot \frac{dy}{dx} = 0

2(y-b) \frac{dy}{dx} = -2(x-a)

\frac{dy}{dx} = -\frac{x-a}{y-b} \quad ... (1)

Next, we find the second derivative,

\frac{d^2y}{dx^2}

, by differentiating

\frac{dy}{dx}

with respect to

x

using the quotient rule.

\frac{d^2y}{dx^2} = -\frac{d}{dx}\left(\frac{x-a}{y-b}\right)

\frac{d^2y}{dx^2} = -\left[ \frac{\frac{d}{dx}(x-a) \cdot (y-b) - (x-a) \cdot \frac{d}{dx}(y-b)}{(y-b)^2} \right]

\frac{d^2y}{dx^2} = -\left[ \frac{1 \cdot (y-b) - (x-a) \cdot \frac{dy}{dx}}{(y-b)^2} \right]

Substitute the expression for

\frac{dy}{dx}

from equation (1):

\frac{d^2y}{dx^2} = -\left[ \frac{y-b - (x-a) \left(-\frac{x-a}{y-b}\right)}{(y-b)^2} \right]

\frac{d^2y}{dx^2} = -\left[ \frac{y-b + \frac{(x-a)^2}{y-b}}{(y-b)^2} \right]

To simplify, multiply the numerator and denominator by

(y-b)

:

\frac{d^2y}{dx^2} = -\left[ \frac{(y-b)^2 + (x-a)^2}{(y-b)^3} \right]

From the original given equation, we know

(x-a)^2 + (y-b)^2 = c^2

. Substituting this into the expression for the second derivative:

\frac{d^2y}{dx^2} = -\frac{c^2}{(y-b)^3} \quad ... (2)

Now we evaluate the expression we need to prove is constant. First, let's compute

1 + \left(\frac{dy}{dx}\right)^2

:

1 + \left(-\frac{x-a}{y-b}\right)^2 = 1 + \frac{(x-a)^2}{(y-b)^2}

= \frac{(y-b)^2 + (x-a)^2}{(y-b)^2}

Using the given equation again, this becomes:

1 + \left(\frac{dy}{dx}\right)^2 = \frac{c^2}{(y-b)^2}

Now substitute this and the second derivative into the required expression:

\frac{\left[1+\left(\frac{dy}{dx}\right)^{2}\right]^{\frac{3}{2}}}{\frac{d^{2} y}{d x^{2}}} = \frac{\left[\frac{c^2}{(y-b)^2}\right]^{\frac{3}{2}}}{-\frac{c^2}{(y-b)^3}}

= \frac{\frac{(c^2)^{\frac{3}{2}}}{((y-b)^2)^{\frac{3}{2}}}}{-\frac{c^2}{(y-b)^3}}

= \frac{\frac{c^3}{|y-b|^3}}{-\frac{c^2}{(y-b)^3}}

= \frac{c^3}{|y-b|^3} \times \frac{-(y-b)^3}{c^2}

Since

|y-b|^3 = \pm (y-b)^3

, the expression becomes:

= \frac{c^3}{\pm (y-b)^3} \times \frac{-(y-b)^3}{c^2} = \mp c

So the value is either

c

or

-c

. Since

c>0

, the expression is a non-zero constant. For example, if we consider the upper semi-circle,

y-b = \sqrt{c^2 - (x-a)^2} > 0

, so

|y-b|=y-b

and the expression is

-c

. For the lower semi-circle,

y-b < 0

, so

|y-b|=-(y-b)

and the expression is again

-c

. Thus, the value of the expression is

-c

, which is a constant and is independent of

a

and

b

.

Final Answer: The expression evaluates to

-c

, which is a constant independent of

a

and

b

. Hence proved.

Q16Miscellaneous Exercise on Chapter 5

If $\cos y=x \cos (a+y)$ , with $\cos a \neq \pm 1$ , prove that $\frac{d y}{d x}=\frac{\cos ^{2}(a+y)}{\sin a}$ .

Solution

To Prove:

\frac{dy}{dx}=\frac{\cos ^{2}(a+y)}{\sin a}

, given

\cos y=x \cos (a+y)

with

\cos a \neq \pm 1

.

Given:

\cos y = x \cos(a+y)

Solution: From the given equation, we can express

x

in terms of

y

:

x = \frac{\cos y}{\cos(a+y)}

Now, we differentiate

x

with respect to

y

using the quotient rule, which is

\frac{d}{dy}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dy} - u \frac{dv}{dy}}{v^2}

. Let

u = \cos y

and

v = \cos(a+y)

. Then

\frac{du}{dy} = -\sin y

and

\frac{dv}{dy} = -\sin(a+y) \cdot \frac{d}{dy}(a+y) = -\sin(a+y)

.

Applying the quotient rule:

\frac{dx}{dy} = \frac{\cos(a+y)(-\sin y) - (\cos y)(-\sin(a+y))}{\cos^2(a+y)}

\frac{dx}{dy} = \frac{-\sin y \cos(a+y) + \cos y \sin(a+y)}{\cos^2(a+y)}

The numerator is of the form

\sin A \cos B - \cos A \sin B = \sin(A-B)

, where

A = a+y

and

B = y

.

\frac{dx}{dy} = \frac{\sin((a+y) - y)}{\cos^2(a+y)}

\frac{dx}{dy} = \frac{\sin a}{\cos^2(a+y)}

We need to find

\frac{dy}{dx}

, which is the reciprocal of

\frac{dx}{dy}

.

\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = \frac{1}{\frac{\sin a}{\cos^2(a+y)}}

\frac{dy}{dx} = \frac{\cos^2(a+y)}{\sin a}

(Note: The condition

\cos a \neq \pm 1

ensures that

\sin a \neq 0

, so the expression is well-defined.)

Hence Proved.

Q17Miscellaneous Exercise on Chapter 5

If $x=a(\cos t+t \sin t)$ and $y=a(\sin t-t \cos t)$ , find $\frac{d^{2} y}{d x^{2}}$ .

Solution

Given: Parametric equations

x = a(\cos t + t \sin t)

y = a(\sin t - t \cos t)

To Find:

\frac{d^2y}{dx^2}

Solution: First, we find

\frac{dx}{dt}

and

\frac{dy}{dt}

.

Differentiating

x

with respect to

t

:

\frac{dx}{dt} = a \frac{d}{dt}(\cos t + t \sin t)

Using the product rule for

t \sin t

:

\frac{dx}{dt} = a(-\sin t + (1 \cdot \sin t + t \cdot \cos t))

\frac{dx}{dt} = a(-\sin t + \sin t + t \cos t)

\frac{dx}{dt} = at \cos t

Differentiating

y

with respect to

t

:

\frac{dy}{dt} = a \frac{d}{dt}(\sin t - t \cos t)

Using the product rule for

t \cos t

:

\frac{dy}{dt} = a(\cos t - (1 \cdot \cos t + t \cdot (-\sin t)))

\frac{dy}{dt} = a(\cos t - (\cos t - t \sin t))

\frac{dy}{dt} = a(\cos t - \cos t + t \sin t)

\frac{dy}{dt} = at \sin t

Now, we find

\frac{dy}{dx}

:

\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{at \sin t}{at \cos t} = \tan t

Next, we find the second derivative,

\frac{d^2y}{dx^2}

.

\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}(\tan t)

Using the chain rule, we differentiate with respect to

t

and multiply by

\frac{dt}{dx}

:

\frac{d^2y}{dx^2} = \frac{d}{dt}(\tan t) \cdot \frac{dt}{dx}

We know that

\frac{d}{dt}(\tan t) = \sec^2 t

and

\frac{dt}{dx} = \frac{1}{dx/dt} = \frac{1}{at \cos t}

.

Substituting these into the expression for the second derivative:

\frac{d^2y}{dx^2} = \sec^2 t \cdot \frac{1}{at \cos t}

Since

\sec t = \frac{1}{\cos t}

, we have

\sec^2 t = \frac{1}{\cos^2 t}

.

\frac{d^2y}{dx^2} = \frac{1}{\cos^2 t} \cdot \frac{1}{at \cos t} = \frac{1}{at \cos^3 t}

This can also be written as

\frac{\sec^3 t}{at}

.

Final Answer:

\frac{d^2y}{dx^2} = \frac{1}{at \cos^3 t}

Q18Miscellaneous Exercise on Chapter 5

If $f(x)=|x|^{3}$ , show that $f^{\prime \prime}(x)$ exists for all real $x$ and find it.

Solution

Given: The function

f(x) = |x|^3

.

To Show:

f''(x)

exists for all real

x

and to find it.

Solution: First, we write the function

f(x)

in a piecewise form. We know that

|x| = x

if

x \geq 0

and

|x| = -x

if

x < 0

. Therefore,

f(x) = (x)^3 = x^3

, if

x \geq 0

.

f(x) = (-x)^3 = -x^3

, if

x < 0

. So, the piecewise definition is:

f(x) = \begin{cases} x^3, & x \geq 0 \ -x^3, & x < 0 \end{cases}

Now, we find the first derivative,

f'(x)

. For

x > 0

,

f'(x) = \frac{d}{dx}(x^3) = 3x^2

. For

x < 0

,

f'(x) = \frac{d}{dx}(-x^3) = -3x^2

.

To check for differentiability at

x=0

, we find the left-hand and right-hand derivatives: LHD at

x=0

:

f'_{-}(0) = \lim_{h \to 0^-} \frac{f(0+h)-f(0)}{h} = \lim_{h \to 0^-} \frac{-h^3 - 0}{h} = \lim_{h \to 0^-} (-h^2) = 0

. RHD at

x=0

:

f'_{+}(0) = \lim_{h \to 0^+} \frac{f(0+h)-f(0)}{h} = \lim_{h \to 0^+} \frac{h^3 - 0}{h} = \lim_{h \to 0^+} (h^2) = 0

. Since LHD = RHD = 0,

f'(0)

exists and

f'(0)=0

. So,

f(x)

is differentiable for all

x \in \mathbb{R}

.

The first derivative is:

f'(x) = \begin{cases} 3x^2, & x \geq 0 \ -3x^2, & x < 0 \end{cases}

Next, we find the second derivative,

f''(x)

. For

x > 0

,

f''(x) = \frac{d}{dx}(3x^2) = 6x

. For

x < 0

,

f''(x) = \frac{d}{dx}(-3x^2) = -6x

.

To check for the existence of

f''(0)

, we find the left-hand and right-hand derivatives of

f'(x)

at

x=0

: LHD of

f'(x)

at

x=0

:

f''_{-}(0) = \lim_{h \to 0^-} \frac{f'(0+h)-f'(0)}{h} = \lim_{h \to 0^-} \frac{-3h^2 - 0}{h} = \lim_{h \to 0^-} (-3h) = 0

. RHD of

f'(x)

at

x=0

:

f''_{+}(0) = \lim_{h \to 0^+} \frac{f'(0+h)-f'(0)}{h} = \lim_{h \to 0^+} \frac{3h^2 - 0}{h} = \lim_{h \to 0^+} (3h) = 0

. Since the left-hand and right-hand derivatives of

f'(x)

at

x=0

are equal,

f''(0)

exists and

f''(0)=0

.

Thus,

f''(x)

exists for all real

x

. The function for

f''(x)

is:

f''(x) = \begin{cases} 6x, & x \geq 0 \ -6x, & x < 0 \end{cases}

This can be written compactly using the absolute value function. Since

|x| = x

for

x \geq 0

and

|x| = -x

for

x < 0

, we have

6x = 6|x|

for

x \geq 0

and

-6x = 6(-x) = 6|x|

for

x < 0

.

Final Answer:

f''(x)

exists for all real

x

and is given by

f''(x) = 6|x|

.

Q19Miscellaneous Exercise on Chapter 5

Using the fact that $\sin (\mathrm{A}+\mathrm{B})=\sin \mathrm{A} \cos \mathrm{B}+\cos \mathrm{A} \sin \mathrm{B}$ and the differentiation, obtain the sum formula for cosines.

Solution

Given: The trigonometric identity

\sin(A+B) = \sin A \cos B + \cos A \sin B

.

To Find: The sum formula for cosines by using differentiation.

Solution: Let us treat

A

as a variable, say

x

, and keep

B

as a constant. The identity becomes:

\sin(x+B) = \sin x \cos B + \cos x \sin B

Now, we differentiate both sides of this equation with respect to

x

. Since

B

is treated as a constant,

\cos B

and

\sin B

are also constants.

Differentiating the Left-Hand Side (LHS) with respect to

x

:

\frac{d}{dx}(\sin(x+B)) = \cos(x+B) \cdot \frac{d}{dx}(x+B)

= \cos(x+B) \cdot (1+0) = \cos(x+B)

Differentiating the Right-Hand Side (RHS) with respect to

x

:

\frac{d}{dx}(\sin x \cos B + \cos x \sin B) = \frac{d}{dx}(\sin x \cos B) + \frac{d}{dx}(\cos x \sin B)

= \cos B \cdot \frac{d}{dx}(\sin x) + \sin B \cdot \frac{d}{dx}(\cos x)

= \cos B (\cos x) + \sin B (-\sin x)

= \cos x \cos B - \sin x \sin B

Equating the differentiated LHS and RHS:

\cos(x+B) = \cos x \cos B - \sin x \sin B

This is the sum formula for cosines. Replacing the variable

x

back with

A

, we get the formula in terms of

A

and

B

.

Final Answer: The sum formula for cosines is obtained as:

\cos(A+B) = \cos A \cos B - \sin A \sin B

Q20Miscellaneous Exercise on Chapter 5

Does there exist a function which is continuous everywhere but not differentiable at exactly two points? Justify your answer.

Solution

Question: Does there exist a function which is continuous everywhere but not differentiable at exactly two points? Justify your answer.

Answer: Yes, such a function exists.

Justification: We can construct such a function using the absolute value function, which is known to be continuous everywhere but not differentiable at the point where its argument is zero.

Consider the function

f(x) = |x-a| + |x-b|

, where

a

and

b

are distinct real numbers. The absolute value function is continuous everywhere. The sum of two continuous functions is also continuous. Therefore,

f(x)

is continuous for all

x \in \mathbb{R}

.

The points where differentiability might fail are

x=a

and

x=b

. Let's choose specific values for

a

and

b

, for example,

a=1

and

b=-1

.

Let the function be

f(x) = |x-1| + |x+1|

.

We can write this as a piecewise function:

For $x < -1$ : $x-1 < 0$ and $x+1 < 0$ . $f(x) = -(x-1) - (x+1) = -x+1-x-1 = -2x$ .
For $-1 \leq x < 1$ : $x-1 < 0$ and $x+1 \geq 0$ . $f(x) = -(x-1) + (x+1) = -x+1+x+1 = 2$ .
For $x \geq 1$ : $x-1 \geq 0$ and $x+1 > 0$ . $f(x) = (x-1) + (x+1) = 2x$ .

So, the function is:

f(x) = \begin{cases} -2x, & \text{if } x < -1 \ 2, & \text{if } -1 \leq x < 1 \ 2x, & \text{if } x \geq 1 \end{cases}

Now, let's check for differentiability at

x=-1

and

x=1

.

At

x=-1

: Left-Hand Derivative (LHD) =

\lim_{h \to 0^-} \frac{f(-1+h)-f(-1)}{h} = \lim_{h \to 0^-} \frac{-2(-1+h) - 2}{h} = \lim_{h \to 0^-} \frac{2-2h-2}{h} = \lim_{h \to 0^-} (-2) = -2

. Right-Hand Derivative (RHD) =

\lim_{h \to 0^+} \frac{f(-1+h)-f(-1)}{h} = \lim_{h \to 0^+} \frac{2 - 2}{h} = \lim_{h \to 0^+} 0 = 0

. Since LHD

\neq

RHD,

f(x)

is not differentiable at

x=-1

.

At

x=1

: Left-Hand Derivative (LHD) =

\lim_{h \to 0^-} \frac{f(1+h)-f(1)}{h} = \lim_{h \to 0^-} \frac{2 - 2}{h} = \lim_{h \to 0^-} 0 = 0

. Right-Hand Derivative (RHD) =

\lim_{h \to 0^+} \frac{f(1+h)-f(1)}{h} = \lim_{h \to 0^+} \frac{2(1+h) - 2}{h} = \lim_{h \to 0^+} \frac{2+2h-2}{h} = \lim_{h \to 0^+} 2 = 2

. Since LHD

\neq

RHD,

f(x)

is not differentiable at

x=1

.

For any other point

x \neq -1

and

x \neq 1

, the function is a simple polynomial (

y=-2x

,

y=2

, or

y=2x

), which is differentiable everywhere in its domain.

Therefore, the function

f(x) = |x-1| + |x+1|

is continuous everywhere but is not differentiable at exactly two points,

x=-1

and

x=1

.

Q21Miscellaneous Exercise on Chapter 5

If $y=\left|\begin{array}{ccc}f(x) & g(x) & h(x) \ l & m & n \ a & b & c\end{array}\right|$ , prove that $\frac{d y}{d x}=\left|\begin{array}{ccc}f^{\prime}(x) & g^{\prime}(x) & h^{\prime}(x) \ l & m & n \ a & b & c\end{array}\right|$

Solution

To Prove: If

y=\left|\begin{array}{ccc}f(x) & g(x) & h(x) \ l & m & n \ a & b & c\end{array}\right|

, then

\frac{d y}{d x}=\left|\begin{array}{ccc}f^{\prime}(x) & g^{\prime}(x) & h^{\prime}(x) \ l & m & n \ a & b & c\end{array}\right|

.

Proof: We are given the function

y

as a determinant:

y=\left|\begin{array}{ccc}f(x) & g(x) & h(x) \ l & m & n \ a & b & c\end{array}\right|

To find the derivative of

y

with respect to

x

, we can first expand the determinant along the first row (R1).

y = f(x) \left|\begin{array}{cc}m & n \ b & c\end{array}\right| - g(x) \left|\begin{array}{cc}l & n \ a & c\end{array}\right| + h(x) \left|\begin{array}{cc}l & m \ a & b\end{array}\right|

y = f(x)(mc - nb) - g(x)(lc - na) + h(x)(lb - ma)

Here,

l, m, n, a, b, c

are constants. Therefore, the cofactors

(mc - nb)

,

(lc - na)

, and

(lb - ma)

are also constants.

Now, we differentiate

y

with respect to

x

using the sum rule and the constant multiple rule for differentiation:

\frac{dy}{dx} = \frac{d}{dx} [f(x)(mc - nb) - g(x)(lc - na) + h(x)(lb - ma)]

\frac{dy}{dx} = (mc - nb) \frac{d}{dx}(f(x)) - (lc - na) \frac{d}{dx}(g(x)) + (lb - ma) \frac{d}{dx}(h(x))

\frac{dy}{dx} = (mc - nb)f'(x) - (lc - na)g'(x) + (lb - ma)h'(x)

Now, let us consider the determinant on the right-hand side of the equation we want to prove:

RHS = \left|\begin{array}{ccc}f^{\prime}(x) & g^{\prime}(x) & h^{\prime}(x) \ l & m & n \ a & b & c\end{array}\right|

Expanding this determinant along the first row (R1):

RHS = f'(x) \left|\begin{array}{cc}m & n \ b & c\end{array}\right| - g'(x) \left|\begin{array}{cc}l & n \ a & c\end{array}\right| + h'(x) \left|\begin{array}{cc}l & m \ a & b\end{array}\right|

RHS = f'(x)(mc - nb) - g'(x)(lc - na) + h'(x)(lb - ma)

Comparing our expressions for

\frac{dy}{dx}

and the RHS, we see that they are identical.

\frac{dy}{dx} = f'(x)(mc - nb) - g'(x)(lc - na) + h'(x)(lb - ma) = RHS

Hence Proved.

Alternative Method: The derivative of a determinant is the sum of the determinants formed by differentiating one row (or column) at a time, keeping other rows (or columns) unchanged. Given

y = D(x) = \left|\begin{array}{ccc}R_1 \ R_2 \ R_3\end{array}\right|

, where

R_1, R_2, R_3

are the rows. Then

\frac{dy}{dx} = \left|\begin{array}{ccc}R_1' \ R_2 \ R_3\end{array}\right| + \left|\begin{array}{ccc}R_1 \ R_2' \ R_3\end{array}\right| + \left|\begin{array}{ccc}R_1 \ R_2 \ R_3'\end{array}\right|

. In our case,

R_1 = (f(x), g(x), h(x))

,

R_2 = (l, m, n)

, and

R_3 = (a, b, c)

.

R_1' = (f'(x), g'(x), h'(x))

. Since

R_2

and

R_3

consist of constants, their derivatives are zero vectors:

R_2' = (0, 0, 0)

and

R_3' = (0, 0, 0)

.

\frac{dy}{dx} = \left|\begin{array}{ccc}f'(x) & g'(x) & h'(x) \ l & m & n \ a & b & c\end{array}\right| + \left|\begin{array}{ccc}f(x) & g(x) & h(x) \ 0 & 0 & 0 \ a & b & c\end{array}\right| + \left|\begin{array}{ccc}f(x) & g(x) & h(x) \ l & m & n \ 0 & 0 & 0\end{array}\right|

A determinant with a row of all zeros is equal to zero.

\frac{dy}{dx} = \left|\begin{array}{ccc}f'(x) & g'(x) & h'(x) \ l & m & n \ a & b & c\end{array}\right| + 0 + 0

\frac{dy}{dx} = \left|\begin{array}{ccc}f'(x) & g'(x) & h'(x) \ l & m & n \ a & b & c\end{array}\right|

Hence Proved.

Q22Miscellaneous Exercise on Chapter 5

If $y=e^{a \cos ^{-1} x},-1 \leq x \leq 1$ , show that $\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}-a^{2} y=0$ .

Solution

Given: The function

y = e^{a \cos^{-1} x}

, for

-1 \leq x \leq 1

.

To Show:

(1-x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} - a^2y = 0

.

Solution: We are given the function:

y = e^{a \cos^{-1} x} \quad ...(1)

First, we differentiate

y

with respect to

x

using the chain rule.

\frac{dy}{dx} = \frac{d}{dx} (e^{a \cos^{-1} x})

\frac{dy}{dx} = e^{a \cos^{-1} x} \cdot \frac{d}{dx}(a \cos^{-1} x)

We know that

\frac{d}{dx}(\cos^{-1} x) = -\frac{1}{\sqrt{1-x^2}}

.

\frac{dy}{dx} = e^{a \cos^{-1} x} \cdot a \left( -\frac{1}{\sqrt{1-x^2}} \right)

Substituting

y = e^{a \cos^{-1} x}

from equation (1):

\frac{dy}{dx} = -\frac{ay}{\sqrt{1-x^2}}

To simplify for the next differentiation, we rearrange the equation:

\sqrt{1-x^2} \frac{dy}{dx} = -ay

Now, square both sides to eliminate the square root:

(1-x^2) \left(\frac{dy}{dx}\right)^2 = (-ay)^2

(1-x^2) \left(\frac{dy}{dx}\right)^2 = a^2y^2 \quad ...(2)

Next, we differentiate equation (2) with respect to

x

. We use the product rule on the left side and the chain rule on the right side.

For the LHS, let

u = 1-x^2

and

v = (\frac{dy}{dx})^2

. Then

u' = -2x

and

v' = 2(\frac{dy}{dx}) \frac{d^2y}{dx^2}

.

\frac{d}{dx} \left[ (1-x^2) \left(\frac{dy}{dx}\right)^2 \right] = \frac{d}{dx}(1-x^2) \cdot \left(\frac{dy}{dx}\right)^2 + (1-x^2) \cdot \frac{d}{dx}\left(\frac{dy}{dx}\right)^2

= -2x \left(\frac{dy}{dx}\right)^2 + (1-x^2) \left( 2 \frac{dy}{dx} \frac{d^2y}{dx^2} \right)

For the RHS:

\frac{d}{dx}(a^2y^2) = a^2 \cdot 2y \frac{dy}{dx}

Equating the derivatives of LHS and RHS:

-2x \left(\frac{dy}{dx}\right)^2 + 2(1-x^2) \frac{dy}{dx} \frac{d^2y}{dx^2} = 2a^2 y \frac{dy}{dx}

Assuming

\frac{dy}{dx} \neq 0

, we can divide the entire equation by

2\frac{dy}{dx}

:

-x \frac{dy}{dx} + (1-x^2) \frac{d^2y}{dx^2} = a^2 y

Rearranging the terms to match the required expression:

(1-x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} - a^2y = 0

Hence Shown.

Continuity and DifferentiabilityClass 12 Mathematics NCERT Solutions

Prove that the function f(x)=5x−3f(x)=5 x-3f(x)=5x−3 is continuous at x=0x=0x=0, at x=−3x=-3x=−3 and at x=5x=5x=5.

Solution

Examine the continuity of the function f(x)=2x2−1f(x)=2 x^{2}-1f(x)=2x2−1 at x=3x=3x=3.

Solution

Examine the following functions for continuity.

Solution

Prove that the function f(x)=xnf(x)=x^{n}f(x)=xn is continuous at x=nx=nx=n, where nnn is a positive integer.

Solution

Is the function fff defined by x, \text { if } x \leq 1 \ 5, \text { if } x>1 \end{array}\right.$$ continuous at $x=0$ ? At $x=1$ ? At $x=2$ ?

Solution

Find all points of discontinuity of fff, where fff is defined by 6. f(x)={2x+3, if x≤2 2x−3, if x>2f(x)=\left\{\begin{array}{l} 2 x+3, \text { if } x \leq 2 \ 2 x-3, \text { if } x>2 \end{array}\right.f(x)={2x+3, if x≤2 2x−3, if x>2​

Solution

Solution

Find all points of discontinuity of fff, where fff is defined by 8. f(x)={∣x∣x, if x≠0 0, if x=0f(x)=\left\{\begin{array}{cc} \frac{|x|}{x}, & \text { if } x \neq 0 \ 0, & \text { if } x=0 \end{array}\right.f(x)={x∣x∣​,​ if x=0 0,​ if x=0​

Solution

Find all points of discontinuity of fff, where fff is defined by 9. f(x)={x∣x∣, if x<0 −1, if x≥0f(x)= \begin{cases}\frac{x}{|x|}, & \text { if } x<0 \ -1, & \text { if } x \geq 0\end{cases}f(x)={∣x∣x​,​ if x<0 −1,​ if x≥0​

Solution

Find all points of discontinuity of fff, where fff is defined by 10. f(x)={x+1, if x≥1 x2+1, if x<1f(x)=\left\{\begin{array}{l} x+1, \text { if } x \geq 1 \ x^{2}+1, \text { if } x<1 \end{array}\right.f(x)={x+1, if x≥1 x2+1, if x<1​

Solution

Find all points of discontinuity of fff, where fff is defined by 11. f(x)={x3−3, if x≤2 x2+1, if x>2f(x)= \begin{cases}x^{3}-3, & \text { if } x \leq 2 \ x^{2}+1, & \text { if } x>2\end{cases}f(x)={x3−3,​ if x≤2 x2+1,​ if x>2​

Solution

Find all points of discontinuity of fff, where fff is defined by 12. f(x)={x10−1, if x≤1 x2, if x>1f(x)= \begin{cases}x^{10}-1, & \text { if } x \leq 1 \ x^{2}, & \text { if } x>1\end{cases}f(x)={x10−1,​ if x≤1 x2,​ if x>1​

Solution

Is the function defined by f(x)={x+5, if x≤1 x−5, if x>1f(x)= \begin{cases}x+5, & \text { if } x \leq 1 \ x-5, & \text { if } x>1\end{cases}f(x)={x+5,​ if x≤1 x−5,​ if x>1​ a continuous function?

Solution

Solution

Discuss the continuity of the function fff, where fff is defined by f(x)={2x, if x<0 0, if 0≤x≤1 4x, if x>1f(x)= \begin{cases}2 x, & \text { if } x<0 \ 0, & \text { if } 0 \leq x \leq 1 \ 4 x, & \text { if } x>1\end{cases}f(x)={2x,​ if x<0 0,​ if 0≤x≤1 4x,​ if x>1​

Solution

Discuss the continuity of the function fff, where fff is defined by f(x)={−2, if x≤−1 2x, if −11f(x)= \begin{cases}-2, & \text { if } x \leq -1 \ 2 x, & \text { if }-11\end{cases}f(x)={−2,​ if x≤−1 2x,​ if −11​

Solution

Find the relationship between aaa and bbb so that the function fff defined by f(x)={ax+1, if x≤3 bx+3, if x>3f(x)= \begin{cases}a x+1, & \text { if } x \leq 3 \ b x+3, & \text { if } x>3\end{cases}f(x)={ax+1,​ if x≤3 bx+3,​ if x>3​ is continuous at x=3x=3x=3.

Solution

Solution

Show that the function defined by g(x)=x−[x]g(x)=x-[x]g(x)=x−[x] is discontinuous at all integral points. Here [x][x][x] denotes the greatest integer less than or equal to xxx.

Solution

Is the function defined by f(x)=x2−sin⁡x+5f(x)=x^{2}-\sin x+5f(x)=x2−sinx+5 continuous at x=πx=\pix=π ?

Solution

Discuss the continuity of the following functions:

Solution

Discuss the continuity of the cosine, cosecant, secant and cotangent functions.

Solution

Find all points of discontinuity of fff, where f(x)={sin⁡xx, if x<0 x+1, if x≥0f(x)= \begin{cases}\frac{\sin x}{x}, & \text { if } x<0 \ x+1, & \text { if } x \geq 0\end{cases}f(x)={xsinx​,​ if x<0 x+1,​ if x≥0​

Solution

Determine if fff defined by f(x)={x2sin⁡1x, if x≠0 0, if x=0f(x)= \begin{cases}x^{2} \sin \frac{1}{x}, & \text { if } x \neq 0 \ 0, & \text { if } x=0\end{cases}f(x)={x2sinx1​,​ if x=0 0,​ if x=0​ is a continuous function?

Solution

Examine the continuity of fff, where fff is defined by f(x)={sin⁡x−cos⁡x, if x≠0 −1, if x=0f(x)= \begin{cases}\sin x-\cos x, & \text { if } x \neq 0 \ -1, & \text { if } x=0\end{cases}f(x)={sinx−cosx,​ if x=0 −1,​ if x=0​

Solution

Solution

Solution

Solution

Solution

Solution

Show that the function defined by f(x)=cos⁡(x2)f(x)=\cos \left(x^{2}\right)f(x)=cos(x2) is a continuous function.

Solution

Show that the function defined by f(x)=∣cos⁡x∣f(x)=|\cos x|f(x)=∣cosx∣ is a continuous function.

Solution

Examine that sin⁡∣x∣\sin |x|sin∣x∣ is a continuous function.

Solution

Find all the points of discontinuity of fff defined by f(x)=∣x∣−∣x+1∣f(x)=|x|-|x+1|f(x)=∣x∣−∣x+1∣.

Solution

Differentiate the functions with respect to xxx in Exercises 1 to 8 . sin⁡(x2+5)\sin \left(x^{2}+5\right)sin(x2+5)

Solution

Differentiate the functions with respect to xxx in Exercises 1 to 8 . 2. cos⁡(sin⁡x)\cos (\sin x)cos(sinx)

Solution

sin⁡(ax+b)\sin (a x+b)sin(ax+b)

Solution

sec⁡(tan⁡(x))\sec (\tan (\sqrt{x}))sec(tan(x​))

Solution

sin⁡(ax+b)cos⁡(cx+d)\frac{\sin (a x+b)}{\cos (c x+d)}cos(cx+d)sin(ax+b)​

Solution

cos⁡x3⋅sin⁡2(x5)\cos x^{3} \cdot \sin ^{2}\left(x^{5}\right)cosx3⋅sin2(x5)

Solution

2cot⁡(x2)2 \sqrt{\cot \left(x^{2}\right)}2cot(x2)​

Solution

cos⁡(x)\cos (\sqrt{x})cos(x​)

Solution

Prove that the function fff given by f(x)=∣x−1∣,x∈Rf(x)=|x-1|, x \in \mathbf{R}f(x)=∣x−1∣,x∈R is not differentiable at x=1x=1x=1.

Solution

Prove that the greatest integer function defined by f(x)=[x],0<x<3f(x)=[x], 0<x<3f(x)=[x],0<x<3 is not differentiable at x=1x=1x=1 and x=2x=2x=2.

Prove that the function $f(x)=5 x-3$ is continuous at $x=0$ , at $x=-3$ and at $x=5$ .

Examine the continuity of the function $f(x)=2 x^{2}-1$ at $x=3$ .

Prove that the function $f(x)=x^{n}$ is continuous at $x=n$ , where $n$ is a positive integer.

Is the function $f$ defined by $x, \text { if } x \leq 1 \ 5, \text { if } x>1 \end{array}\right.$$ continuous at $x=0$ ? At $x=1$ ? At $x=2$ ?$

Find all points of discontinuity of $f$ , where $f$ is defined by 6. $f(x)=\left\{\begin{array}{l} 2 x+3, \text { if } x \leq 2 \ 2 x-3, \text { if } x>2 \end{array}\right.$

Find all points of discontinuity of $f$ , where $f$ is defined by 8. $f(x)=\left\{\begin{array}{cc} \frac{|x|}{x}, & \text { if } x \neq 0 \ 0, & \text { if } x=0 \end{array}\right.$

Find all points of discontinuity of $f$ , where $f$ is defined by 9. $f(x)= \begin{cases}\frac{x}{|x|}, & \text { if } x<0 \ -1, & \text { if } x \geq 0\end{cases}$

Find all points of discontinuity of $f$ , where $f$ is defined by 10. $f(x)=\left\{\begin{array}{l} x+1, \text { if } x \geq 1 \ x^{2}+1, \text { if } x<1 \end{array}\right.$

Find all points of discontinuity of $f$ , where $f$ is defined by 11. $f(x)= \begin{cases}x^{3}-3, & \text { if } x \leq 2 \ x^{2}+1, & \text { if } x>2\end{cases}$

Find all points of discontinuity of $f$ , where $f$ is defined by 12. $f(x)= \begin{cases}x^{10}-1, & \text { if } x \leq 1 \ x^{2}, & \text { if } x>1\end{cases}$

Is the function defined by $f(x)= \begin{cases}x+5, & \text { if } x \leq 1 \ x-5, & \text { if } x>1\end{cases}$ a continuous function?

Discuss the continuity of the function $f$ , where $f$ is defined by $f(x)= \begin{cases}2 x, & \text { if } x<0 \ 0, & \text { if } 0 \leq x \leq 1 \ 4 x, & \text { if } x>1\end{cases}$

Discuss the continuity of the function $f$ , where $f$ is defined by $f(x)= \begin{cases}-2, & \text { if } x \leq -1 \ 2 x, & \text { if }-11\end{cases}$

Find the relationship between $a$ and $b$ so that the function $f$ defined by $f(x)= \begin{cases}a x+1, & \text { if } x \leq 3 \ b x+3, & \text { if } x>3\end{cases}$ is continuous at $x=3$ .

Show that the function defined by $g(x)=x-[x]$ is discontinuous at all integral points. Here $[x]$ denotes the greatest integer less than or equal to $x$ .

Is the function defined by $f(x)=x^{2}-\sin x+5$ continuous at $x=\pi$ ?

Find all points of discontinuity of $f$ , where $f(x)= \begin{cases}\frac{\sin x}{x}, & \text { if } x<0 \ x+1, & \text { if } x \geq 0\end{cases}$

Determine if $f$ defined by $f(x)= \begin{cases}x^{2} \sin \frac{1}{x}, & \text { if } x \neq 0 \ 0, & \text { if } x=0\end{cases}$ is a continuous function?

Examine the continuity of $f$ , where $f$ is defined by $f(x)= \begin{cases}\sin x-\cos x, & \text { if } x \neq 0 \ -1, & \text { if } x=0\end{cases}$

Show that the function defined by $f(x)=\cos \left(x^{2}\right)$ is a continuous function.

Show that the function defined by $f(x)=|\cos x|$ is a continuous function.

Examine that $\sin |x|$ is a continuous function.

Find all the points of discontinuity of $f$ defined by $f(x)=|x|-|x+1|$ .

Differentiate the functions with respect to $x$ in Exercises 1 to 8 . $\sin \left(x^{2}+5\right)$

Differentiate the functions with respect to $x$ in Exercises 1 to 8 . 2. $\cos (\sin x)$

$\sin (a x+b)$

$\sec (\tan (\sqrt{x}))$

$\frac{\sin (a x+b)}{\cos (c x+d)}$

$\cos x^{3} \cdot \sin ^{2}\left(x^{5}\right)$

$2 \sqrt{\cot \left(x^{2}\right)}$

$\cos (\sqrt{x})$

Prove that the function $f$ given by $f(x)=|x-1|, x \in \mathbf{R}$ is not differentiable at $x=1$ .

Prove that the greatest integer function defined by $f(x)=[x], 0<x<3$ is not differentiable at $x=1$ and $x=2$ .

$2 x+3 y=\sin x$

$2 x+3 y=\sin y$

$a x+b y^{2}=\cos y$

$x y+y^{2}=\tan x+y$

$x^{2}+x y+y^{2}=100$

$x^{3}+x^{2} y+x y^{2}+y^{3}=81$

$\sin ^{2} y+\cos x y=\kappa$

$\sin ^{2} x+\cos ^{2} y=1$

$y=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$

$y=\tan ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right),-\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}$

$y=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right), 0<x<1$

$y=\sin ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right), 0<x<1$

$y=\cos ^{-1}\left(\frac{2 x}{1+x^{2}}\right),-1<x<1$

$y=\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right),-\frac{1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}$

$y=\sec ^{-1}\left(\frac{1}{2 x^{2}-1}\right), 0<x<\frac{1}{\sqrt{2}}$

$\frac{e^{x}}{\sin x}$

$e^{\sin ^{-1} x}$

$e^{x^{3}}$

$\sin (\tan ^{-1} e^{-x})$

$\log (\cos e^{x})$

$e^{x}+e^{x^{2}}+\ldots+e^{x^{5}}$

$\sqrt{e^{\sqrt{x}}}, x>0$

$\log (\log x), x>1$

$\frac{\cos x}{\log x}, x>0$

$\cos (\log x+e^{x}), x>0$

$\cos x \cdot \cos 2 x \cdot \cos 3 x$

$\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}$

$(\log x)^{\cos x}$

Differentiate the functions given in Exercises 1 to 11 w.r.t. $x$ . 4. $x^{x}-2^{\sin x}$

Differentiate the functions given in Exercises 1 to 11 w.r.t. $x$ . 5. $(x+3)^{2}\cdot(x+4)^{3}\cdot(x+5)^{4}$

Differentiate the functions given in Exercises 1 to 11 w.r.t. $x$ . 6. $\left(x+\frac{1}{x}\right)^{x}+x^{\left(1+\frac{1}{x}\right)}$