# Derivative Examples

Consider a function which involves the change in velocity of a vehicle moving from one point to another. The change in velocity is certainly dependent on the speed and direction in which the vehicle is travelling. If the acceleration is to be calculated, then the limits of the function are essential. The theory of derivative is derived from limits. This article deals with the concept of derivatives along with a few solved derivative examples.

## Derivative Definition

A function which denotes the rate of change of the other function can be called the derivative of that function. The method of finding a derivative can be called differentiation. It is represented as a dependent variable in terms of an independent variable through an equation. The derivative is a value which changes with respect to its input. A derivative is denoted as f’(x).

## Derivative Formula

The change in the slope of a function can be termed as derivative. It is represented as follows:

$slope=\frac{change\ in\ x}{change\ in\ y}=\frac{\Delta y}{\Delta x}=\frac{dy}{dx}=\frac{f(x+\Delta x)-f(x)}{\Delta x}$

### Related topics

Differentiation of functions

Methods of differentiation

## Derivative Rules

Below are some important rules used to solve the derivative problems.

Constant rule: $\frac{d}{dx}[C] = 0$

Power rule: $\frac{d}{dx}[x^{n}] = n \; x^{n-1}$

Constant multiple rule: $\frac{d}{dx}[c.u] = c.\frac{du}{dx}$

Sum and difference rule: $\frac{d}{dx}[u\pm v] = \frac{du}{dx}\pm \frac{dv}{dx}$

Product rule: $\frac{d}{dx}[u. v] =u. \frac{dv}{dx}+v.\frac{du}{dx}$

Quotient rule: $\frac{d}{dx}[\frac{u}{v}]=\frac{v.\frac{du}{dx}-u.\frac{dv}{dx}}{v^{2}}$

## Some basic derivatives

$\frac{d}{dx}[x] = 1\\\frac{d}{{dx}}\sqrt x= \frac{1}{{2\sqrt x }}\\\frac{d}{{dx}}\sqrt {f(x)} = \frac{1}{{2\sqrt {f(x)} }}\frac{d}{{dx}}f(x) = \frac{1}{{2\sqrt {f(x)} }}f'(x)\\\frac{d}{{dx}}c \cdot f(x) = c\frac{d}{{dx}}f(x) = c \cdot f'(x)\\\frac{d}{{dx}}\ln x = \frac{1}{x}\\\frac{d}{{dx}}{\log _a}x = \frac{1}{{x\ln a}}\\\frac{d}{{dx}}\ln f(x) = \frac{1}{{f(x)}}\frac{d}{{dx}}f(x)\\\frac{d}{{dx}}{\log _a}f(x) = \frac{1}{{f(x)\ln a}}\frac{d}{{dx}}f(x)\\\frac{d}{{dx}}{e^x} = {e^x}\\\frac{d}{{dx}}{e^{f(x)}} = {e^{f(x)}}\frac{d}{{dx}}f(x)\\\frac{d}{{dx}}{a^x} = {a^x}\ln a\\\frac{d}{{dx}}{a^{f(x)}} = {a^{f(x)}}\ln a\frac{d}{{dx}}f(x)\\\frac{d}{{dx}}{x^x} = {x^x}(1 + \ln x)\\\frac{d}{{dx}}sinx = cosx\\\frac{d}{dx}cosx = -sinx\\\frac{d}{{dx}}tanx = sec{^2}x\\\frac{d}{{dx}}cotx = – cosec{^2}x\\\frac{d}{{dx}}secx = secx \cdot tanx\\\frac{d}{{dx}}cosec x = – co\sec x \cdot cotx\\\frac{d}{{dx}}sinhx = coshx\\\frac{d}{{dx}}coshx = sinhx\\\frac{d}{{dx}}tanhx = sec{h^2}x\\\frac{d}{{dx}}cothx =- cosec {h^2}x\\\frac{d}{{dx}}sechx =- sechx \cdot tanhx\\\frac{d}{{dx}}cosec hx =- co\sec hx \cdot cothx\\\frac{d}{dx}sin^{-1}x=\frac{1}{\sqrt{1-x^{2}}}\\\frac{d}{dx}cos^{-1}x=\frac{-1}{\sqrt {1–x^{2}}}\\\frac{d}{dx}tan^{-1}x = \frac{1}{1+{x^2}}\\$

## Examples on Derivative for IIT JEE

Example 1: Find the derivative of x6+x3+2

Solution:

Using the power rule (d/dx)x6 = 6x5

(d/dx)x3 = 3x2

(d/dx)2 = 0

Hence the derivative of x6+x3+2 = 6x5+3x2

Example 2: Solve $\frac{d}{dx}tan^{-1}\frac{cosx}{1+sinx}$

Solution:

$\frac{d}{dx}tan^{-1}\frac{cosx}{1+sinx}$

Example 3: What is the differential coefficient of ax + log x . sinx?

Solution:

Let y = ax + log x . sinx

Differentiating w.r.t. x, we get

dy / dx = ax loge(a) + (1 / x) sin x + log x . cos x

Example 4: If f(x) = logx (logx), then f′(x) at x = e is _____.

Solution:

f(x) = logx (logx)

= log (log x) / logx

f′(x)= 1 / x − 1 / x log (logx) / (logx)2

⇒ f′(e) = [1 / e − 0] / 1

= 1 / e

Example 5: What is the local maximum value of the function log x / x ?

Solution:

Let f(x) = log x / x

⇒f′(x) = 1 / x2 − log x / x2

For maximum or minimum value of f (x), Find  f′(x) = 0

f′(x) = 1 − logex / x2 = 0 or 1− logex / x2 = 0

∴ logex = 1

or x = e, which lie in (0,∞).

For x = e,

d2y / dx2 =−1 / e3, which is −ve.

Hence y is maximum at x = e and its maximum value = loge / e = 1 / e.

Example 6: If f(2) = 4, f′(2) = 1 then what is the value of limx→2 x f(2) − 2 f(x) / x−2?

Solution:

Given f(2) = 4, f′(2) = 1

limx→2 x f(2) − 2 f(x) / (x−2) = limx→2 x f(2) − 2 f(2) + 2 f(2) − 2 f(x) / (x−2)

[limx→2 (x−2) f(2) / (x − 2)] − [limx→2 2 f(x) − 2 f(2) / (x−2)]

f(2) − 2 limx→2 f(x) − f(2) / [x−2] = f(2) − 2 f′(2)

= 4− 2(1)

= 4−2

= 2

Example 7: In the Mean-Value theorem f(b) − f(a) / b−a = f′(c), if a = 0, b = 1 / 2 and f(x) = x (x−1) (x−2) , then what is the value of c?

Solution:

From mean value theorem f′(c)= f(b) − f(a) / b−a

Given that a=0, f(a)=0 and b = 1 / 2, f(b) = 3 / 8

f′(x) = (x−1) (x−2) + x (x−2) + x (x−1)

f′(c) = (c−1) (c−2) + c (c−2) + c (c−1)

= c2 − 3c +2 + c2 − 2c + c2 − c

f′(c) = 3c2 −6c + 2

According to mean value theorem,

f′(c)= f(b) − f(a) / b−a

3c2 − 6c + 2= (3/8)−0 / (1/2) − 0 = 3 / 4

3c2 − 6c + 5 / 4 = 0

On solving the above equation the value of c obtained is

c = 6 ± √[36−15] / 2 × 3

= [6 ± √21] / 6

c = [1 ± √21] / 6

Example 8: Find the derivative of function f given by f(x) = (2x2-3)(x4+3x-1)

Solution:

Here f(x) is the product of two functions.

u = 2x2-3

v = x4+3x-1

And  f(x) = uv

So we use the product rule to differentiate f(x).

f’(x) = u’v+uv’

u’ = 4x

v’ = 4x3+3

f’(x) = 4x+4x3+3 = 4x3+4x+3

Hence the derivative of the function is 4x3+4x+3.