DIFFERENTIATION USING SUBSITUTION

Example 1 :

Differentiate sin^-1 (3x - 4x³) with respect to x

Solution :

Let y  =  sin^-1 (3x - 4x³) 

First let us consider (3x - 4x³).

If we plug sin θ instead of x in the given function we will get

3 sin θ - 4 sin³ θ

This the formula for sin 3θ.

So, put x  =  sin θ and θ  =  sin^-1 x

y = sin^-1 [3(sin θ) - 4(sin θ)³]

y = sin^-1 [3 sin θ - 4 sin³ θ]

y = sin^-1 [sin 3θ]

y = 3 θ

y = 3 sin^-1x

differentiating with respect to x on both sides

dy/dx  =  3(1/√(1 - x²))

dy/dx  =  3/√(1 -  x²)

Example 2 :

Differentiate tan^-1 [(1+x²)/(1-x²)] with respect to x

Solution : 

Take y = tan^-1[(1+x²)/(1-x²)]

tan 2θ = (1 - tan²θ) / (1 + tan²θ)

then x = tan θ

y = tan^-1[(1 - tan²θ) / (1 + tan²θ)]

y = tan^-1[tan 2θ]

y = 2θ

Here θ = tan^-1 x

y = 2tan^-1 x

dy/dx = 2 (1/1+x²)

= 2/(1+x²)

Example 3 :

Differentiate the following

sin^-1[2x/(1+x²)]

Solution :

Let y  =  sin^-1[2x/(1+x²)]

Let x  =  tan a

a = tan^-1x

y = sin^-1[2tan a/(1+tan² a)]

y = sin^-1[sin 2a]

y = 2a

y = 2 tan^-1x

Differentiate with respect to x, we get

dy/dx = 2(1/(1+x²))

dy/dx = 2/(1+x²)

Example 4 :

Differentiate the following

tan^-1[√(1 - cos x)/(1+cosx)]

Solution :

Let y  =  tan^-1[√(1 - cos x)/(1+cosx)]

Trigonometric formula for (1-cos x)/(1+cosx)  =  tan²(x/2)

By applying trigonometric formula, we may find derivatives easily. 

 y  =  tan^-1[√tan²(x/2)]

y = x/2

Differentiating with respect to "x"

dy/dx  =  1/2

Example 5 :

Differentiate the following

tan^-1[6x/1-9x²]

Solution :

Let y  =  tan^-1[6x/1-9x²]

y  =  tan^-1[2(3x) /1-(3x)²]

Let 3x = tan θ

y  =  tan^-1[2 tan θ /1-tan²θ]

2 tan θ /1-tan²θ  = tan 2θ

y  =  tan^-1[tan 2θ]

y = 2θ

From 3x = tan θ, θ = tan^-1(3x)

Applying the value of θ, we get

y = 2tan^-1(3x)

Differentiating with respect to x, we get

dy/dx = 2(1/1 + (3x)²) (3)

= 6/(1 + 9x²)

Example 6 :

Differentiate the following

cos (2tan^-1[√(1-x)/(1+x)])

Solution :

Let y  =  cos (2tan^-1[√(1-x)/(1+x)])

let x = cos θ

Differentiating with respect to x, we get

dx/dθ  =  -sin θ

 y  =  cos (2tan^-1[√(1-cosθ)/(1+cosθ)])

 y  =  cos (2tan^-1[√tan²(θ/2])

 y  =  cos (2tan^-1[tan(θ/2])

 y  =  cos θ

From x = cos θ, θ = cos^-1(x)

y = cos (cos^-1(x))

y = x

Differentiating with respect to x, we get

dy/dx = 1

Example 7 :

If y = tan^-1(1 + x)/(1 - x), then find y'

Solution :

Let

tan π/4 = 1

tan θ = x

(1 + x)/(1 - x) = (tan π/4 + tan θ) / (1 - tan π/4 tan θ)

tan (A + B) = (tan A + tan B) / (1 - tan A tan B)

= tan (π/4 + θ)

y = tan^-1[tan (π/4 + θ)]

y = π/4 + θ

Applying the value of θ as tan^-1(x)

y = π/4 + tan^-1(x)

Differentiate with respect to x, we get

dy/dx = 0 + 1/(1 + x²)

dy/dx = 1/(1 + x²)

Example 8 :

Find f'(x) if f(x) = cos^-1(4x³ - 3x)

Solution :

f(x) = cos^-1(4x³ - 3x)

Let x = cos θ

= cos^-1(4(cos θ)³ - 3 cos θ)

= cos^-1(4cos³ θ - 3 cos θ)

= cos^-1(cos 3θ)

f(x) = 3θ

Applying the value of θ from x = cos θ

θ = cos ^-1(x)

f(x) = 3 cos ^-1(x)

f'(x) = 3 (-1/√1-x²)

f'(x) = (-3/√1-x²)

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