Mathematics

Session

Differentiation - 2

Session Objectives

Fundamental Rules, Product Rule and Quotient Rule

Differentiation of Function of a Function

Differentiation by Trigonometric Substitutions

Differentiation of Implicit Functions

Class Exercise

Fundamental Rules

d d2 cƒ x = c ƒ x

dx dx

d d d3 f x ± g x = f x ± g x

dx dx dx

d1 c = 0

dx

Example-1

Differentiate the following:

x1tanx+2sinx+3cosx - logx - e

2

Solution

x1Let y = tanx+2sinx+3cosx - logx - e

2

Differentiating y with respect to x, we get

xdy d 1= tanx+2sinx+3cosx - logx - e

dx dx 2

xdy d d d 1 d d= tanx +2 sinx +3 cosx - logx - e

dx dx dx dx 2 dx dx

2 xdy 1=sec x+2cosx - 3sinx - - e

dx 2x

Product Rule

If f x and g x are differentiable functions, then

d d df x .g x = f x . g x + f x .g x

dx dx dx

Example-2

Differentiating w.r.t. x, we get

2 2dy d d= x sinx logx +logx x sinx

dx dx dx

Differentiate: w.r.t. x.2x sinxlogx

2 2 21x

d d= x sinx× +logx x sinx +sinx x

dx dx

2= xsinx+x cosxlogx+2xsinxlogx

Solution

Let y =x2sinxlogx

Quotient Rule

If f x and g x are differentiable functions and g x 0 , then

2

d dg x × f x - f x × g xf xd dx dx=

dx g x g x

Example-3

Differentiate: w.r.t. x.1+logx1- logx

1+logxLet y =

1- logx

Solution:

Differentiating w.r.t. x, we get

2

d d1- logx 1+logx - 1+logx 1- logxdy dx dx=

dx 1- logx

2

1 11- logx × - 1+logx × -

x x=

1- logx

21- logx+1+logx

=x 1- logx 2

2=

x 1- logx

Differentiation of Function of a Function

dy dy dt= ×

dx dt dx

y = ƒ t t =g xNote: If and

, then

d d dfog x = fog x × g x

dx dg x dx

If ƒ(x) and g(x) are differentiable functions, then ƒog is also differentiable

(Chain Rule)

Example-4

Differentiate log log logx w. r. t. x.

Solution:

Let y =log log logx

Differentiating y w.r.t. x, we get

dy 1 d= × log(logx)

dx log(logx) dx

dy 1 1 d= × × (logx)

dx log(logx) logx dx

dy 1 1 1=

dx log(logx) logx x

dy 1=

dx xlogxlog(logx)

Example-5

23 2

dy5xIf y = + sin (2x + 3), find .

dx1 - x

23 2

5xSolution: We have y = +sin (2x +3)

1- x

1

-2 23y =5x 1- x +sin (2x +3)

1 4

- -2 23 3dy 1=5 1- x +x - 1- x (-2x) +2sin(2x +3)cos(2x +3)×2

dx 3

Continued

2

1 42 23 3

dy 5 10x= + +4sin(2x +3)cos(2x +3)

dx1- x 3 1- x

2 2

42 3

15 1- x +10xdy= +2sin(4x +6) [ 2sinxcosx = sin2x]

dx3 1- x

2

42 3

5 3- xdy= +2sin(4x +6)

dx3 1- x

Trigonometric Substitutions

2 2a +x x =atanθ or acotθ

2 2a - x x =asinθ or acosθ

2 2x - a x =asecθ or acosecθ

a+x a- xor

a- x a+xx =acos2θ

2 2 2 2

2 2 2 2a +x a - x

ora - x a +x

2 2x =a cos2θ

Example-62 2

-1

2 2

1+x + 1- x dyIF y = tan , find

dx1+x - 1- x

2 2-1

2 2

1+x + 1- xWe have y = tan

1+x - 1- x

Putting x2 = cos2

-1 1+cos2θ + 1- cos2θy = tan

1+cos2θ - 1- cos2θ

Solution:

2 2-1

2 2

2cos θ + 2sin θy = tan

2cos θ - 2sin θ

-1 cosθ+sinθy = tan

cosθ- sinθ

Continued

-1 1+tanθy = tan

1- tanθ

-1y = tan tan +θ4

y = +θ4

-1 21y = + cos x

4 2

Differentiating y w.r.t. x, we get

-1 2dy d 1 d= + cos x

dx dx 4 2 dx

22

dy 1 -1=0+ 2x

dx 21- x

4

dy -x=

dx 1- x

Example-7

22 2

2

dy 1- yf 1- x + 1- y = a(x - y), prove that =

dx 1- xI

2 2Solution: We have 1- x + 1- y = a(x - y)

Putting x = sin and y = sin

2 21- sin + 1- sin = a(sin - sin )

cos +cos = a(sin - sin )

cos +cosa=

sin - sin

Continued+ -

2cos cos2 2

a=+ -

2cos sin2 2

-a= cot

2

-1 -1 -1sin x - sin y = 2cot a

-1 -1 -1d dsin x - sin y = 2cot a

dx dx

2 2

dy1 1- = 0

dx1- x 1- y

-1- = 2cot a

2

2

1- ydy=

dx 1- x

Differentiation of Implicit Functions

ƒ x, y =0

y is not expressible directly in terms of x

Example-8

3 3 dyIf xy - yx = x, find

dx

We have xy3 – yx3 = x

3 3d d dxy - yx = x

dx dx dx

2 3 2 3dy dyx 3y +y 1- y 3x +x =1

dx dx

Solution:

Solution Cont.

2 3 2 3dy dy3xy +y - 3x y - x =1

dx dx

2 2 2 2dyx 3y - x +y y - 3x =1

dx

2 2 2 2dyx 3y - x =1- y y - 3x

dx

2 2 3 2

2 32 2

1- y y - 3xdy 1- y +3x y= =

dx 3xy - xx 3y - x

Thank you