Differentiation

Have you ever inflated a spherical balloon?

What do we observe? As we start inflating the balloon its radius starts increasing and consequently, its volume also starts increasing.

https://goo.gl/images/vrhn3e

Volume V of the spherical balloon is given by, V= $\frac{4}{3}$ $\Pi r^{2}$ .

If r=1 cm then V = $\frac{4}{3}$ $\Pi cm^{3}$ .

If r=2 cm then V= $\frac{32}{3}$ $\Pi cm^{3}$ .

If we want to find the rate at which the volume V increases as the radius r increases, then we want to find the rate of change of Volume with respect to the radius.

So, if we want to find the rate of change of a dependent variable (Volume V here) with respect to independent variable (radius r here), we want to find $\frac{dV}{dr}$ or $V^{{1}}$ (r).

Here we will introduce the concept of Differentiation.

I. Derivative of a function at a point.

Let f(x) be a real valued function and x = c be any point in its domain, then the derivative of f(x) at x= c is denoted by $f^{{1}}$ (c) and it is defined by f(c)=lim $\left ( \frac{f(c+h)-f(c)}{h} \right )$ provided this limit exists.

II. Derivative of a function in general.

In general, the derivative of f(x) is denoted by $f^{{1}}$ (x) and it is defined by

$f^{{'}}$ (x)=lim $\left ( \frac{f(c+h)-f(c)}{h} \right )$ provided this limit exists.

Also, if y = f(x), then the derivative or the differential coefficient of f(x) with respect to x is denoted by $\frac{dy}{dx}$ or $y^{{'}}$ or $f^{{'}}$ (x).

The process of finding the derivative of a function is called Differentiation.

To differentiate a function $f^{{'}}$ (x) means to find its derivative $f^{{'}}$ (x).

Derivative of some standard functions

Algebra of Derivative of functions

If f(x) and g(x) be any two functions such that their derivatives are defined over the common domain then

$\frac{d}{dx}$ [f(x)+g(x)]= $\frac{d}{dx}$ f(x)+ $\frac{d}{dx}$ g(x)
$\frac{d}{dx}$ [f(x)-g(x)]= $\frac{d}{dx}$ f(x) – $\frac{d}{dx}$ g(x)
$\frac{d}{dx}$ [f(x)g(x)]=g(x) $\frac{d}{dx}$ f(x) + f(x) $\frac{d}{dx}$ g(x)

This is also known as Product Rule of Differentiation.

$\frac{d}{dx}$ $\left [ \frac{f(x)}{g(x)} \right ]$ = $\frac{g(x)\frac{d}{dx}f(x)-f(x)\frac{d}{dx}g(x)}{\left [ g(x) \right ]^2}$ where g(x) $\neq$ 0.

This is also known as Quotient Rule of Differentiation

Examples

Now let’s consider some examples on differentiation .

Example 1: Find the derivative of f(x)= $x^{{11}}$ + 3x.

$f^{{1}}$ (x)= $\frac{d}{dx}$ ( $x^{{11}}$ ) +3 $\frac{d}{dx}$ (x)=11 $x^{{10}}$ +3(1)=11 $x^{{10}}$ +3

Example 2: Find the derivative of f(x)=7Sin x-3Tan x.

$f^{{1}}$ (x)=7 $\frac{d}{dx}$ (Sin x) -3 $\frac{d}{dx}$ (Tan x)=7Cos x – 3 $Sec^{{2}}$ x

Example 3: Find the derivative of f(x)=Sin xCos x .

$f^{{1}}$ (x)= $\frac{d}{dx}$ (Sin xCos x)

Cos x $\frac{d}{dx}$ (Sin x) + Sin x $\frac{d}{dx}$ (Cos x)= Cos x(Cos x) + Sin x(-Sin x)using Product rule.

= $Cos^{{2}}$ x – $Sin^{{2}}$ x

Example 4: Find the derivative of f(x)= $\frac{Sin x}{Cos x}$ .

$f^{{'}}$ (x)= $\frac{d}{dx}$ $\left ( \frac{Sin x}{Cos x} \right )$

= $\frac{Cosx\frac{d}{dx}(Sinx)-Sinx\frac{d}{dx}(Cosx)}{Cos^2x}$ = $\frac{Cos x(Cos x)-Sin x(-Sin x)}{Cos^2x}$ using Quotient Rule.

= $\frac{Cos^2x+Sin^2x}{Cos^2x}$ = $\frac{1}{Cos^2x}$

Example 5: Find the derivative of f(x)= $e^x$ +20 $x^12$ +15.

$f^1$ (x)= $\frac{d}{dx}$ ( $e^x$ )+20 $\frac{d}{dx}$ ( $x^12$ )+ $\frac{d}{dx}$ (15)=

= $e^{{x}}$ +20(12 $x^{{11}}$ )+0= $e^x$ +240 $x^{{11}}$ .

Check Point

Find the derivative of the following functions f(x) with respect to x.

f(x)= $x^{15}$ +20x
f(x)=Cosx -3Tan x
f(x)=Sinx Tanx
f(x)= $\frac{Cosx}{Sinx}$
f(x)=5 $e^{{x}}$ + 2 $x^{{17}}$ + 27

Answer Key

15 $x^{{4}}$ + 20
-Sinx – 3 $Sec^{{2}}$ x
Cosx(Tan x)+Sin x( $Sec^{{2}}$ x)
$\frac{-1}{Sin^2 x}$
5 $e^x$ + 34 $x^16$

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