# Differentiation of Functions

Differentiation is a method to find rates of change. Need for differentiation of a function arises quite often in mathematical, engineering and scientific problems. If the function has a closed-form representation in terms of standard calculus, then its derivative can be found exactly. However, in many situations, we may not know the exact function. What we know is only the values of the function at a discrete set of points. This section focuses on learning calculus and based on differentiation of functions of one variable.

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## Derivative of a function

The derivative 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 is often written using “dy over dx” (meaning the difference in y divided by the difference in x).

The derivative can be expressed as: ${\displaystyle {\frac {dy}{dx}}}$.

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

In mathematical terms,

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

That is, as the distance between the two x points (h) becomes closer to zero, the slope of the line between them comes closer to resembling a tangent line.

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

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

### 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 g^n(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: $y={{e}^{{{\tan }^{-1}}x}}\Rightarrow \frac{dy}{dx}=\frac{{{d}^{{{\tan }^{-1}}}}x}{1+{{x}^{2}}}$ $\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{(1-2x){{e}^{ta{{n}^{-1}}x}}}{{{(1+{{x}^{2}})}^{2}}}$

Again,

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