# DIFFERENTIATION BY TRIGONOMETRIC SUBSTITUTION

Example 1 :

Differentiate the following

cos-1(1 -x2)/(1+x2)

Solution :

Let y  =  cos-1(1 -x2)/(1+x2)

x = tan θ

y  =  cos-1(1- tan2 θ)/(1+tan2 θ)

y  =  cos-1(cos 2θ)

y = 2θ

Differentiate with respect to x, we get

dy/dx  =  2 (dθ/dx)    ------(1)

x = tan θ

1  =  sec2θ (dθ/dx)

dθ/dx  =  1/sec2θ

By applying the value of dθ/dx in (1), we get

dy/dx  =  2 (1/sec2θ)

=  2/(1 + tan2θ)

=  2/(1 + x2)

Example 2 :

Differentiate the following

sin-1(3x - 4x3)

Solution :

Let y  =  sin-1(3x - 4x3)

x = sin θ

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 = 3θ

Differentiate with respect to x, we get

dy/dx  =  3 (dθ/dx)    ------(1)

x = sin θ

1  =  cos θ (dθ/dx)

dθ/dx  =  1/cosθ

By applying the value of dθ/dx in (1), we get

dy/dx  =  3 (1/cos sec2θ)

=  2/(1 + tan2θ)

=  2/(1 + x2)

Example 3 :

Differentiate the following

tan-1[(cos x + sin x) / (cos x - sin x)]

Solution :

Let y  =  tan-1[(cos x + sin x) / (cos x - sin x)]

Divide every terms inside the parentheses by cos x, we get

y  =  tan-1[(1 + tan x) / (1 - tan x)]

y  =  tan-1[(tan π/4) + tan x) / (1 (tan π/4) tan x)]

y  =  tan-1[tan ((π/4) + x)]

y  =  (π/4) + x

Differentiate with respect to "x", we get

dy/dx  =  1

Example 4 :

Find the derivative of sin x2 with respect to x2

Solution :

Let y = sin x2

here we have to differentiate with respect to "x2" not "x"

dy/dx=  cosx2

Incase we differentiate with respect to "x", we get

dy/dx  =  cosx2 (2x)

=  2x cosx2

Example 5 :

Find the derivative of sin-1(2x / (1 + x2)) with respect to tan-1 x

Solution :

Let y  =  sin-1(2x / (1 + x2))

x = tan θ

y  =  sin-1(2tan θ / (1 + tan2 θ))

y  =  sin-1(2tan θ / (1 + tan2 θ))

y  =  sin-1(sin 2θ)

y = 2θ

y = 2 tan-1x

dy/dx  =  2 [1/(1 + x2)]

dy/dx  =  2/(1 + x2)

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