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)]

Let t  =   (1+x²)/(1-x²)

So the function has become y = tan^-1 t

To differentiate this function with respect to x we have to write the formula required.

dy/dx  =  (dy/du) x (dt/dx)

t  =  [ (1+x²)/(1-x²) ]

For differentiating this we have to apply the quotient rule. So u  =  1+x² and v  =  1-x²

u' = 0 + 2x             v' = 0 - 2x

u' = 2x                  v' = -2x

(U/V)' =  [VU' - UV'] /V²

dt/dx  =  [( 1-x²) 2x - (1+x²)(-2x)]/(1-x²)²

dt/dx  =  [(2x - 2x³) - (-2x - 2x³)]/(1-x²)²

dt/dx  =  [(2x - 2x³ + 2x + 2x³)]/(1-x²)²

dt/dx  =  [4x/(1-x²)²]

y  =  tan^-1 t

dy/dt  =  1/(1+t²)

dy/dt  =  1/(1+[(1+x²)/(1-x²)]²)

dy/dt  =  1/(1+[(1+x²)²/(1-x²)²])

dy/dt  =  1/[(1-x²)]²+[(1+x²)²]/(1-x²)]²)

dy/dt  =  (1-x²)]²/[(1-x²)]²+[(1+x²)²]

dy/dt  =  (1-x²)]²/[(1 + x⁴ - 2x²)+(1+ x⁴ + 2x²)]

dy/dt  =  (1-x²)²/(2 + 2x⁴)

dy/dx  =  (dy/dt) x  (dt/dx)

=  (1-x²)²/(2 + 2x⁴) x  4x/(1-x²)²

=  4x/(2 + 2x⁴)

=  4x/2(1+x⁴)

dy/dx  =  2x/(1+x⁴)

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[sin2a]

y  =  2tan^-1x

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

dy/dx  =  2 (dθ/dx)  ------(1)

3x  =  tan θ

3(1)  =  sec²θ(dθ/dx)

3/sec²θ  =  (dθ/dx)

Applying dθ/dx  =  3/sec²θ in (1)

dy/dx  =  2 (3/sec²θ) 

   =  2(3)/(1  + tan²θ)

   =  6/(1 + (3x)²)

   =  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

Differentiating with respect to θ.

dy/dx  =  -sin θ (dθ/dx)    -----(1)

Now let us apply the value of dx/dθ  =  -sin θ in (1)

dy/dx  =  -sin θ (-1/sinθ) 

dy/dx  =  1

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