**Function and the inverse-function:**

A function f(x) is said to have its inverse say g(x) [ where g(x) = f^{-1} (x)] , if and only if the f(x) is a one- to one function.

If f(x) and g(x) are inverse functions to each other, then the following conditions are true.

- g(f(x))= x
- f(g(x))=x

**Horizontal line test:**

** **

If a function passes the horizontal line test, then it is a one-to-one function. Else it is not a one-to-one function and we can not have the inverse to the function.

**The relation between the derivative of f(x)and the derivative of f**^{-1}(x) :

^{-1}(x) :

** **

We have the derivative of a function f(x) = f ’(x)

and the derivative of inverse function f^{-1}(x) is represented by [ f^{-1}(x) ]’

Let us derive the relation between these two derivatives.

let us assume that f^{-1}(x)= g(x).

Then as per the inverse properties, we can say that g(f(x)) =x

Let us derivate both sides

**Therefore, the derivative of the inverse of a function is 1 over the derivative of the function.**

**The derivative of an inverse function Worked out examples.**

** **

**Example1:**

** **

**What is the value of the d/dx [ f**^{-1}(x)] when x=2,given that f(x)= x^{3}+x

^{-1}(x)] when x=2,given that f(x)= x

^{3}+x

**And f**^{-1}(2)=1.

^{-1}(2)=1.

** **__Solution:__

Given that f(x) = x^{3}+x

Therefore the f’ (x) = 3x^{2}+1

Given that the f^{-1}(2)=1 .Therefore f(1) =2.

We have the formula

**To find:**

d/dx [ f^{-1}(x)] when x=2

we initially assumed that the f^{-1}(x) = g(x)

hence d/dx [ f^{-1}(x)] =g’(x).

Therefore we need to find the g’(x) at x=2.

That is the value of g’(2).

From the formula

On comparision we can conclude that f(x) =2

And we can observe that f(x) =x^{3} +x can be 2, if and only if x=1.

Therefore,

F’(x)= 3x^{2}+1 .Therefore f’(1) = 3(1)^{2}+1 =4

**Conclusion :**

g’(2) = 1/f’(1) = 1/4

Hence, the d/dx [ f^{-1}(x)] when x=2 is** 1/4.**

The derivative of an inverse function at x=2 is 1/4.