Derivatives of a function are the slope of the tangent at an arbitrary point of the curve y = f (x). Derivatives are very much useful in our every day life, like finding the rate of volume of water flowing in a cylindrical tank, when the rate at which the height of water level is increasing is given and vice-versa. Also we are aware that the area of a circle depends on its radius. Therefore, the rate of increase in area depends on the rate of increase in the radius. The functions of graph will be either explicit or implicit function. Hence we should know to find the derivative of the given function to find its slope, rate of change etc. In this section let us see how to find the derivative of a given function and some examples related to this.

## Derivative Rules

Let us see some of the rules to find the derivatives of a function.
Let f(x) and g (x) be the two functions where x $\epsilon$ R.

### Sum Rule:

The derivative of sum of the functions, is the sum of derivative of the derivatives of each function.
(i. e), $\frac{d}{dx}$ [ f (x) + g (x) ] = $\frac{d}{dx}$ [ f (x) ] + $\frac{d}{dx}$ [ g (x) ]

In general, $\frac{d}{dx}$ [ u + v + w + ............... ] = $\frac{du}{dx}$ + $\frac{dv}{dx}$ + $\frac{dw}{dx}$ + . . . . . . . . . . . . . .

### Difference Rule:

The derivative of difference of the functions, is the differenceof derivative of the derivatives of each function.
(i. e), $\frac{d}{dx}$ [ f (x) - g (x) ] = $\frac{d}{dx}$ [ f (x) ] - $\frac{d}{dx}$ [ g (x) ]

In general, $\frac{d}{dx}$ [ u - v - w - ............... ] = $\frac{du}{dx}$ - $\frac{dv}{dx}$ - $\frac{dw}{dx}$ - . . . . . . . . . . . .

### Product Rule:

The derivative of the product of the functions is found by the following method.
$\frac{d}{dx}$ [ f (x). g (x) ] = f (x) . $\frac{d}{dx}$[ g (x) ] + g (x) . $\frac{d}{dx}$ [ f (x) ]

The above can also be stated as
$\frac{d}{dx}$ [ u. v ] = u . $\frac{dv}{dx}$ + v. $\frac{du}{dx}$

The above rule can be extended to the product of any finite number of functions, as follows.
$\frac{d}{dx}$ [ u. v. w ] = v. w. $\frac{du}{dx}$ + u . w . $\frac{dv}{dx}$ + u . v $\frac{dw}{dx}$

### Quotient Rule:

If u and v are functions of x and v (x) $\ne$ 0, then
$\frac{\mathrm{d} }{\mathrm{d} x}\left (\frac{u}{v} \right)$ = $\frac{v.\frac{\mathrm{du}}{\mathrm{d} x}-u.\frac{\mathrm{dv} }{\mathrm{d} x} }{v^{2}}$

### Chain Rule - Differentiation of Composite functions:

In the case of composite function, where y = (f o g) (x), then
$\frac{dy}{dx}$ = $\frac{d}{dx}$ [ (f o g) (x) ] = f ' [ g (x) ] g ' (x)

Also, if y = f (u) , u = g (v) and v = h (x), then
$\frac{dy}{du}$ = f ' (u),

$\frac{du}{dv}$ = g ' (v)
and $\frac{dv}{dx}$ = h ' (x)
Therefore, $\frac{dy}{dx}$ = $\frac{dy}{du}$ . $\frac{du}{dv}$ . $\frac{dv}{dx}$

= f ' (u) . g ' (v) . h ' (x)

= f ' [ g { h (x) } ] . g ' { h (x) } . h ' (x)

## Derivative Rules Examples

### Solved Examples

Question 1: Sum Rule: Differentiate f (x ) = x3 + 4 x2 + 3 x + 7
Solution:

We have f (x) = x3 + 4 x2 + 3 x + 7
Therefore using the rule,

$\frac{d}{dx}$ [ xn ] = n . xn-1

we get f' (x) = $\frac{d}{dx}$ [ x3 + 4 x2 + 3 x + 7 ]

= $\frac{d}{dx}$ [ x3 ] + 4 . $\frac{d}{dx}$ [ x2 ] + 3 $\frac{d}{dx}$ [ x ] + $\frac{d}{dx}$ [ 7 ]

= 3 x2 + 4 (2 x) + 3 (1) + 0
f' (x) = 3 x2 + 8 x + 3

Question 2: Difference rule: Differentiate, f (x) = ex - $\frac{3}{x^{2}}$
Solution:

We have, f (x) = ex - $\frac{3}{x^{2}}$

using difference rule, we get,

f' (x) = $\frac{d}{dx}$ [ ex - $\frac{3}{x^{2}}$ ]

= $\frac{d}{dx}$ [ ex - 3 . x-2 ]

$\frac{d}{dx}$ [ ex ] - $\frac{d}{dx}$ [ 3 . x-2 ] = ex - 3 (- 2) x-2-1

= ex + 6 . x-3

f' (x) = ex + $\frac{6}{x^{3}}$

Question 3: Product Rule: Differentiate f (x) = (x2 + 4x + 5) (x3 + 5 x)
Solution:

We have f (x) = (x2 + 4x + 5) (x3 + 5 x)
According to product rule
(u v) ' = u . v' + u' . v
Let u = x2 + 4x + 5 , v = x3 + 5 x
Therefore, u' = 2x + 4 and v' = 3 x2 + 5
Therefore, f' (x) = u . v' + u' . v
= (x2 + 4x + 5)(3 x2 + 5) + (x3 + 5 x) (2x + 4) .
= x2 (3 x2 + 5) + 4 x (3 x2 + 5) + 5 (3 x2 + 5) + x3 (2x + 4) + 5x (2x + 4)
= 3 x4 + 5 x2 + 12 x3 + 20 x + 15 x2 + 25 + 2 x4 + 4 x3 + 10 x2 + 20 x
= 3 x4 + 2 x4 + 12 x3 + 4 x3 + 5 x2 + 15 x2 + 10 x2 + 20 x + 20 x + 25
f' (x) = 5 x4 + 16 x3 + 30 x2 + 40 x + 25

Question 4: Quotient Rule: Differentiate f (x) = $\frac{(2x+3)}{(x^{2}-5)}$ with respect to x.
Solution:

We have f (x) = $\frac{(2x+3)}{(x^{2}-5)}$

According to Quotient rule,

($\frac{u}{v}$)' = $\frac{v\;u'\;-\;v\;u'}{v^{2}}$

From the given function f (x), we have

u = (2x + 3) and v = (x2 - 5)

Therefore, u' = 2 and v' = 2x

Substituting in the rule we get,

f' (x) = $\frac{(x^{2}-5)(2)\;-\;(2x+3)(2x)}{(x^{2}\;-\;5)^{2}}$

= $\frac{(2x^{2}-10)\;-\;(4x^{2}+6x)}{(x^{2}\;-\;5)^{2}}$

= $\frac{2x^{2}-10\;-\;4x^{2}-6x}{(x^{2}\;-\;5)^{2}}$

= $\frac{-2x^{2}-10\;-6x}{(x^{2}\;-\;5)^{2}}$

f' (x) = - 2 [ $\frac{(x^{2}+3x+5)}{(x^{2}\;-\;5)^{2}}$ ]

Question 5: Chain Rule: Differentiate f (x) = (3 x - x3 + 4 x2)5
Solution:

According to chain rule,
The above function can be written as
f (u) = u5 , where u = 3 x - x3 + 4 x2
Therefore, f' (u) = 5 u4 and u' = 3 - 3 x2 + 8 x
=> f' (x) = f' (u). u' (x)
= 5 u4 . (3 - 3 x2 + 8 x) [ Substituting u = 3 x - x3 + 4 x2 ]
= 5 (3 x - x3 + 4 x2)4 . (3 - 3 x2 + 8 x)
f' (x) = 5 (3 - 3 x2 + 8 x) . (3 x - x3 + 4 x2)4