 # Differentiation of Functions

Differentiation is a method to find rates of change. It is an important topic for the JEE exam. Derivative of a function y = f(x) of a variable x is the rate of change of y with respect to rate of change of x. This article helps you to learn the derivative of a function, standard derivatives, theorems of derivatives, differentiation of implicit function and higher order derivatives along with solved examples.

## How to Differentiate a Function

The differentiation of a function is a way to show the rate of change of a function at a given point. For real-valued functions, it is the slope of the tangent line at a point on a graph.

The derivative of y with respect to x is defined as the change in y over the change in x, as the distance between $x_{0}}\ and \ x_{1}$ becomes infinitely small (infinitesimal).  The derivative is often written as ${\frac {dy}{dx}}$.

In mathematical terms,

If f is a real valued function and a is a point in its domain of definition. The derivative of f at a is defined by

$f'(a)=\lim _{h\to 0}{\frac {f(a+h)-f(a)}{h}}$

## Standard Derivatives

1] ${dx}\left(u^n\right)=nu^{n-1}\frac{du}{dx}$

2] $\frac{d}{dx}\left(c\right)=0,\; \text{where c is a constant}$

3] $\text{If } y=F(u)\text{ where }u=f(x)\text{ then }\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}$

(Chain Rule or Function of a Function Rule)

4] Derivatives of trigonometric functions

$\frac{d}{dx}\left(\sin u\right)=\cos u\frac{du}{dx}\\ \frac{d}{dx}\left(\cos u\right)=-\sin u\frac{du}{dx}\\ \frac{d}{dx}\left(\tan u\right)=\sec^2 u\frac{du}{dx}\\ \frac{d}{dx}\left(\sec u\right)=\sec u\tan u\frac{du}{dx}\\ \frac{d}{dx}(cosec\ u)=-cosecu\ cotu\ \frac{du}{dx}\\ \frac{d}{dx}(cot\ u)=-cosec^2u\ \frac{du}{dx}\\$

5] Derivatives of inverse trigonometric functions

$\frac{d}{dx}\left(\sin^{-1} u\right)=\frac{1}{\sqrt{1-u^2}}\frac{du}{dx},\qquad -11\\ \frac{d}{dx}\left(\sec^{-1} u\right)=\frac{1}{|u|\sqrt{u^2-1}}\frac{du}{dx}\qquad |u|>1\\ \frac{d}{dx}\left(\cot^{-1} u\right)=-\frac{1}{1+u^2}\frac{du}{dx}\\$

6] Exponential and logarithmic functions
$\frac{d}{dx}\left(e^u\right)=e^u\frac{du}{dx}\\ \frac{d}{dx}\left(a^u\right)=a^u\ln a\frac{du}{dx}, \text{where }a>0, a\ne1\\ \frac{d}{dx}\left(\ln u\right)=\frac{1}{u}\frac{du}{dx}\\ \frac{d}{dx}\left(\ln_a u\right)=\frac{1}{u\ln a}\frac{du}{dx},\text{ where }a>0, a\ne 1\\$

7] Hyperbolic functions

$\frac{d}{dx}\left(\sinh u\right)=\cosh u\frac{du}{dx}\\ \frac{d}{dx}\left(\cosh u\right)=\sinh u\frac{du}{dx}\\ \frac{d}{dx}\left(\tanh u\right)=sech^{2}u\frac{du}{dx}\\ \frac{d}{dx}\ (sech u)=-sechu\ tanhu\frac{du}{dx}\\ \frac{d}{dx}\ (cosech u)=-cosechu\ cothu\frac{du}{dx}\\ \frac{d}{dx}\ (coth u)=-cosech^{2}u\frac{du}{dx}\\$

8] Inverse hyperbolic functions
$\frac{d}{dx}sinh^{-1} u=\frac{1}{\sqrt{1+u^2}}\frac{du}{dx}\\ \frac{d}{dx}cosh^{-1} u=\frac{1}{\sqrt{u^2-1}}\frac{du}{dx},\qquad u>1\\ \frac{d}{dx}tanh^{-1} u=\frac{1}{1-u^2}\frac{du}{dx},\qquad |u|<1\\ \frac{d}{dx}cosech^{-1} u=-\frac{1}{|u|\sqrt{u^2+1}}\frac{du}{dx}\qquad u\ne 0\\ \frac{d}{dx}sech^{-1} u=-\frac{1}{u\sqrt{1-u^2}}\frac{du}{dx}\qquad 01\\$

## Some Standard Substitution

### Expression Substitution

Simple tricks to solve complicated differential equations are listed here:

If function contains $\sqrt{{{a}^{2}}-{{x}^{2}}}$ ; then substitute x = a sinθ or x = a cosθ

If function contains $\sqrt{{{a}^{2}}+{{x}^{2}}}$ ; then substitute x = a cotθ or x = a tan θ

If function contains $\sqrt{{{x}^{2}}+{{a}^{2}}}$ ; then substitute x = a cosecθ or x = a sec θ

## Theorems of Derivatives

Find below some the important theorem results:

$\frac{d}{dx}[u\pm v]=\frac{du}{dx}\pm\frac{dv}{dx}\\ \frac{d}{dx}uv=u\frac{dv}{dx}+v\frac{du}{dx}\qquad\text{(Product Rule)}\\ \frac{d}{dx}\frac{u}{v}=\frac{v\frac{du}{dx}- u\frac{dv}{dx}}{v^2}\qquad\text{(Quotient Rule)}\\\frac{d}{dx}\left( k\,f(x) \right)=k\frac{d}{dx}(f(x)),\,\,where\,k\,is\,constant$

Example 1: Find $\frac{dy}{dx}$for y = x sinx log x

Solution: ${{y}^{‘}}=1\times \sin x\times \log x+x\cos x\log x+x\sin x\times \frac{1}{x}=\sin x\,\log x+x\cos x\log x+\sin x$

Example 2: Find $\frac{dy}{dx}\,\,for\,y=\sin ({{x}^{2}}+1)$

Solution: ${{y}^{‘}}=\cos ({{x}^{2}}+1)\times 2x$

= 2x cos (x2 + 1)

### Differentiation of Implict Function

In implicit differentiation, also known as Chain Rule, differentiate both sides of an equation with two given variables by considering one of the variable as a function of the second variable. In short, differentiate the given function with respect to x and solve for dy/dx. Let us have a look on some the examples.

Example 1: If x2 + 2xy + y3 = 4, find $\frac{dy}{dx}$

Solution: Differentiating both sides w.r.t. x, we get $\frac{d}{dx}({{x}^{2}})+2\frac{d}{dx}(xy)+\frac{d}{dx}({{y}^{3}})=\frac{d}{dx}(4)$ $=2x+2x\frac{dy}{dx}+2y+3{{y}^{2}}.\frac{dy}{dx}=0$ $\Rightarrow \frac{dy}{dx}=\frac{-2(x+y)}{(2x+3{{y}^{2}})}$

Example 2: Differentiable log sin x w.r.t $\sqrt{\cos x}$

Solution: Let u = log sin x  and v = $\sqrt{\cos x}$ $\frac{du}{dx}=\cot x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\And \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{dv}{dx}=\frac{-\sin x}{2\sqrt{\cos x}}$,

$\frac{du}{dx}=\frac{du/dx}{dv/dx}=\frac{\cot x}{\frac{-\sin x}{2\sqrt{\cos x}}}=-2\sqrt{\cos x}\cot x\cos ec(x)$

### Higher order Derivatives

Differentiation process can be continued upto nth derivation of a function. Usually we deal with first order and second order derivatives of the functions.

$\frac{dy}{dx}$ is the first derivative of y w.r.t x

$\frac{d^2y}{dx^2}$ is the second derivative of y w.r.t x

Similarly, finding the third, fourth, fifth and successive derivatives of any function, say g(x), which are known as higher-order derivatives of g(x).
nth order derivative numerical notation is gn(x) or $\frac{d^ny}{dx^n}$

Example: If $y={{e}^{{{\tan }^{-1}}x}},then\,prove\,that\,(1+{{x}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}=(1-2x)\frac{dy}{dx}$

Solution:

$y={{e}^{{{\tan }^{-1}}x}}$

dy/dx = ${{e}^{{{\tan }^{-1}}x}}\frac{1}{1+x^{2}}$

$\frac{e^{tan^{-1}x}}{1+x^{2}}$  …(i)

d2y/dx$\frac{(1+x^{2})e^{tan^{-1}x}\frac{1}{1+x^{2}}-2xe^{tan^{-1}x}}{(1+x^{2})^{2}}$

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

(d2y/dx)(1 + x2) = $\frac{(1-2x)e^{tan^{-1}x}}{(1+x^{2})}$

= (1-2x)dy/dx      (from eqn (i))

Hence proved.