Inverse Function

In this page inverse function we are going to see some example problems to understand this topic much better.

Example 1:

Differentiate  cos⁻¹[ (1-x)/(1+x) ]

Solution

First let us take y = cos⁻¹[ (1-x)/(1+x) ]

                      t = [(1-x)/(1+x)]

  So the given function is becoming y = cos⁻¹ t. Now we have to differentiate [(1-x)/(1+x)] using quotient rule. So let us consider u as (1 -x ) and v as (1 + x).

Formula for quotient rule :

(U/V)' =  [VU' - UV'] /V²


 u = 1 - x                     v = 1 + x

du/dx = 0-1                dv/dx = 0 + 1

du/dx = -1                  dv/dx = 1

 u' = -1                         v' = 1

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

                        = (-1 - x -1 + x) /(1+x)²

                        = -2/(1+x)²

So far we have found the derivative of t with respect to x using quotient rule for differentiation. Now we are going to apply the

Formula for chain rule :

            dy/dx = (dy/dt) (dt/dx)

 y = cos⁻¹ t                        t = [(1-x)/(1+x)]

 dy/dt = -1/(√(1-t²)           dt/dx = -2/(1+x)²

               dy/dx = [-1/(√(1-t²)] x [-2/(1+x)²]

               =  { -1/√(1-[(1-x)/(1+x)]² } x {-2/(1+x)²}

               =  {-1/√1- (1-x)²/(1+x)²} x {-2/(1+x)²}

               = {-1/√(1+x)² - (1-x)²/(1+x)²} x {-2/(1+x)²}

               = {-1/√(1 + x² + 2x) -(1 + x² - 2x) /(1+x)²} x {-2/(1+x)²}

               = {-1/√1 + x² + 2x -1 - x² + 2x) /(1+x)²} x {-2/(1+x)²}

               = {-1/√(4x/(1+x)²)} x {-2/(1+x)²}

               =  {-1/2√(x)/(1+x)} x {-2/(1+x)²}

               = {-(1+x)/2√(x)} x {-2/(1+x)²}

               =  1/√(x)(1+x)


Example 2:

Differentiate sin ⁻¹ (x² + 2x + 5 )

Solution:

Let y = sin ⁻¹ (x² + 2x + 5 )

     u = x² + 2x + 5

Now we can write the given function as y = sin⁻¹ u

Formula for chain rule:

dy/dx = (dy/du) x (du/dx)

    u = x² + 2x + 5                        y = sin⁻¹ u

    du/dx = 2x + 2(1) + 0              dy/du = 1/√(1-u²)

    du/dx = 2x + 2

dy/dx = [1/√(1-u²)] x (2x + 2)

         = [1/√(1-(x² + 2x + 5)²)] x (2x + 2)  

          = 2x + 2/√(1-(x² + 2x + 5)²        





Inverse Function to First Principles