Key Points

A function f(x) is continuous at a point x = c if its Left Hand Limit (LHL), Right Hand Limit (RHL), and the value of the function at that point are equal. Mathematically, $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$.

If two real functions f and g are continuous at x=c, then their sum (f+g), difference (f-g), and product (f.g) are also continuous at x=c. Their quotient (f/g) is continuous at x=c provided g(c) is not equal to 0.

A function f(x) is differentiable at a point x = c if its Left Hand Derivative (LHD) and Right Hand Derivative (RHD) exist and are equal. LHD is $\lim_{h \to 0^-} \frac{f(c+h)-f(c)}{h}$ and RHD is $\lim_{h \to 0^+} \frac{f(c+h)-f(c)}{h}$.

If a function is differentiable at a point, it is necessarily continuous at that point. The converse is not true; a function can be continuous but not differentiable, for example, $f(x) = |x|$ at $x=0$.

The chain rule is used to differentiate composite functions. If $y = f(u)$ and $u = g(x)$, then the derivative of y with respect to x is $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

The derivative of the product of two functions u(x) and v(x) is given by $(uv)' = u'v + uv'$. In Leibniz notation, $\frac{d}{dx}(uv) = v \frac{du}{dx} + u \frac{dv}{dx}$.

The derivative of the quotient of two functions u(x) and v(x) is given by $(\frac{u}{v})' = \frac{u'v - uv'}{v^2}$, provided $v(x) \neq 0$.

Key derivatives are: $\frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}}$, $\frac{d}{dx}(\cos^{-1} x) = -\frac{1}{\sqrt{1-x^2}}$, and $\frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2}$.

The derivative of the natural exponential function is $\frac{d}{dx}(e^x) = e^x$, and for a general base a, $\frac{d}{dx}(a^x) = a^x \log a$. The derivative of the natural logarithm is $\frac{d}{dx}(\log x) = \frac{1}{x}$.

This technique is used for functions of the form $y = [u(x)]^{v(x)}$ or for complex products and quotients. Take the natural logarithm of both sides, simplify using log properties, and then differentiate implicitly.

Used when y cannot be easily expressed as a function of x. Differentiate both sides of the equation with respect to x, treating y as a function of x, and then solve for $\frac{dy}{dx}$.

If $x = f(t)$ and $y = g(t)$, the derivative $\frac{dy}{dx}$ is found by the formula $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$, provided that $\frac{dx}{dt} \neq 0$.

The second order derivative is the derivative of the first derivative. It is denoted by $\frac{d^2y}{dx^2}$, $f''(x)$, or $y''$. It is found by differentiating $\frac{dy}{dx}$ with respect to x.

The fundamental derivatives are: $\frac{d}{dx}(\sin x) = \cos x$, $\frac{d}{dx}(\cos x) = -\sin x$, and $\frac{d}{dx}(\tan x) = \sec^2 x$.