## 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

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

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]

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  If you have any feedback about our math content, please mail us :

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