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 xdx dx

d d d3 f x ± g x = f x ± g xdx dx dx

d1 c = 0dx

Example-1Differentiate the following:

x1tanx+2sinx+3cosx - logx - e2

Solutionx1Let y = tanx+2sinx+3cosx - logx - e2

Differentiating y with respect to x, we get

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

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

2 xdy 1=sec x+2cosx - 3sinx - - edx 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 xdx dx dx

Example-2

Differentiating w.r.t. x, we get

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

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

2 2 21x

d d= x sinx× +logx x sinx +sinx xdx dx

2= xsinx+x cosxlogx+2xsinxlogx

SolutionLet 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-3Differentiate: w.r.t. x.1+logx

1- 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 22=

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 xdx dg x dx

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

(Chain Rule)

Example-4Differentiate 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-52

3 2dy5xIf 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)×2dx 3

Continued

21 4

2 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- xora- x a+xx =acos2θ

2 2 2 22 2 2 2

a +x a - xora - x a +x

2 2x =a cos2θ

Example-62 2-12 2

1+x + 1- x dyIF y = tan , find dx1+x - 1- x

2 2-12 2

1+x + 1- xWe have y = tan1+x - 1- x

Putting x2 = cos2

-1 1+cos2θ + 1- cos2θy = tan1+cos2θ - 1- cos2θ

Solution:

2 2-12 2

2cos θ + 2sin θy = tan2cos 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 x4 2

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

-1 2dy d 1 d= + cos xdx dx 4 2 dx

22

dy 1 -1=0+ 2xdx 2 1- x

4dy -x=dx 1- x

Example-722 22

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 2a= + -2cos sin2 2

-a= cot 2

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

-1 -1 -1d dsin x - sin y = 2cot adx dx

2 2dy1 1- = 0dx1- x 1- y

-1- =2cot a

2

21- ydy =dx 1- x

Differentiation of Implicit Functions

ƒ x, y =0

y is not expressible directly in terms of x

Example-83 3 dyIf xy - yx = x, find dx

We have xy3 – yx3 = x

3 3d d dxy - yx = xdx dx dx

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

Solution:

Solution Cont.

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

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

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

2 2 3 22 32 2

1- y y - 3xdy 1- y +3x y= =dx 3xy - xx 3y - x

Thank you