For a function $f(x)$ the Inverse function $f^{-1} (x)$ reverses or undoes the operation done by the function. This leads to rule on composition that $(fof^{-1})(x)$ = $x$ or $(f^{-1} o f)(x)$ = $x$. That is the composition of a function and its inverse gives the identity. This is what reversing the effect of the function does.  In this lesion we shall take a look into finding the derivative of the inverse functions. We shall take a look with  general function and then find out derivative of some inverse functions.

## How to find Derivative of Inverse Function

Let $f(x)$ be a function with inverse $f^{-1} (x)$ and both $f(x)$ and $f^{-1} (x)$ are differentiable at $x$ and $f' (f^{-1} (x)) \neq 0$, then the derivative of $f^{-1} (x)$ is given by $\frac{d}{dx}$ $f^{-1} (x)$ = $\frac{1}{(f' (f^{-1} (x))}$

Let us see how this is obtained

We know  the composition of the function and its inverse gives the identity function $x$.

$f(f^{-1} (x))$ = $x$

Differentiating with respect to x on both the sides using chain rule we get

$f'(f^{-1}(x))$ $\times$ $\frac{d}{dx}$ $f^{-1}x$ = 1

Divide both sides by  $f'(f^{-1} (x))$

We get

$\frac{d}{dx}$ $f^{-1}(x)$ = $\frac{1}{f'(f^{-1}(x))}$

## Derivative of Inverse Function and logarithms

Under this heading we are going to derive the derivative of inverse trigonometric functions and logarithmic functions
Example 1:

Find the derivative of $y$ = $sin^{(-1)⁡} (x)$
Solution:

$y$ = $sin^{-1}(x)$

Apply sine function on both the sides

$sin⁡(y)$ = $sin⁡(sin^{-1}(x))$

We know the composition of the function and its inverse gives us the identity $x$.

$sin⁡(y)$ = $x$

Differentiating both sides with respect to $x$

$cos(y)$ $\frac{d}{dx}$ = $1$

$\frac{d}{dx}$ = $\frac{1}{cos(y)}$

We know that   $cos^2⁡ y + sin^2⁡ y$ = $1$  which leads to $cos^2⁡ y$ = $1 - sin^2⁡y$  and $cos⁡(y)$ = $\sqrt{(1-sin^2⁡y )}$

$\frac{dy}{dx}$ = $\frac{1}{\sqrt{1 - sin^2(y)}}$

$\frac{dy}{dx}$ = $\frac{1}{\sqrt{1 - x^2}}$

$\frac{d}{dx}$ $sin^{-1}x$ = $\frac{1}{\sqrt{1-x^2}}$
Example 2:

Find the derivative of $y$ = $cos^{-1}x$

Solution:

$y$ = $cos^{-1}x$

Apply cosine on both the sides

$cos⁡(y)$ = $cos(cos^{-1}(x))$

$cos(y)$ = $x$

Differentiating with respect to $x$

$- sin(y)$ $\frac{dy}{dx}$ = $1$

$\frac{dy}{dx}$ = - $\frac{1}{sin(y)}$

We know that $cos^2⁡ y + sin^2⁡ y$ = $1$  which leads to $cos^2⁡ y$ = $1 - cos^2⁡y$  and $cos⁡(y)$ = $\sqrt{(1-cos^2⁡y )}$

$\frac{dy}{dx}$ = - $\frac{1}{\sqrt{1 - cos^2(y)}}$

$\frac{dy}{dx}$ = - $\frac{1}{\sqrt{1 - x^2}}$
Example 3:

Find the derivative of $y$ = $tan^{-1}x$

Solution:

$y$ = $tan^{-1}x$

Applying tangent function on both the sides we get

$tan(y)$ = $x$

Differentiating with respect to $x$

$sec^2(y)$ $\frac{dy}{dx}$ = $1$

$\frac{dy}{dx}$ = $\frac{1}{sec^2(y)}$

$\frac{dy}{dx}$ = $\frac{1}{1 + tan^2(y)}$

$\frac{dy}{dx}$ = $\frac{1}{1+x^2}$

$\frac{d}{dx}$ $tan^{-1}(x)$ = $\frac{1}{1+x^2}$
Example 4:

Find the derivative of $y$ = $csc^{-1}(x)$

Solution:

$y$ = $csc^{-1}(x)$

$csc(y)$ = $x$

Differentiating with respect to $x$

$-csc(y)\ cot(y)$ $\frac{dy}{dx}$  = $1$

$\frac{dy}{dx}$ = $\frac{1}{-csc(y) cot(y)}$

$\frac{dy}{dx}$ = - $\frac{1}{csc(y) cot(y)}$

Use the identity  $1 + cot^2A$ = $csc^2 A$ and rewrite $cot⁡(y)$ = $\sqrt{csc^2(y) - 1}$

$\frac{dy}{dx}$ = - $\frac{1}{csc(y) \sqrt{csc^2(y) - 1}}$

$\frac{dy}{dx}$ = $\frac{1}{x \sqrt{x^2 - 1}}$
Example 5:

Find the derivative of $y$ = $sec^{-1}(x)$

Solution:

$y$ = $sec^{-1}(x)$

$sec(y)$ = $x$

Differentiating with respect to $x$

$sec(y)\ tan(y)$ $\frac{dy}{dx}$ = $1$

$\frac{dy}{dx}$ = $\frac{1}{sec(y) tan(y)}$

Use the identity $1 + tan^2A$ = $sec^2 A$ and rewrite $tan⁡(y)$ = $\sqrt{sec^2(y) - 1}$

$\frac{dy}{dx}$ = $\frac{1}{sec(y) \sqrt{sec^2(y) - 1}}$

$\frac{dy}{dx}$ = $\frac{1}{x \sqrt{x^2 - 1}}$
Example 6:

Find the derivative of $y$ = $cot^{-1}(x)$

Solution:

$y$ = $cot^{-1}(x)$

Applying tangent function on both the sides we get

$cot⁡(y)$ = $x$

Differentiating with respect to $x$

$- csc^2(y)$ $\frac{dy}{dx}$ = $1$

$\frac{dy}{dx}$ = - $\frac{1}{csc^2(y)}$

$\frac{dy}{dx}$ = -$\frac{1}{1+cot^2(y)}$

$\frac{dy}{dx}$ = -$\frac{1}{1+x^2}$

$\frac{d}{dx}$ $cot^{-1}(x)$ = -$\frac{1}{1+x^2}$
Example 7:

Find the derivative of $y$ = $ln(x)$

Solution:

$\frac{dy}{dx}$ = $\frac{1}{x}$
Example 8:

Find the derivative of $y$ = $log(x)$

Solution:

$y$ = $log⁡(x)$

$y$ = $\frac{ln(x)}{ln(10)}$

$\frac{dy}{dx}$ = $\frac{1}{ln(10)}$ $\times$ $\frac{1}{x}$

$\frac{dy}{dx}$ = $\frac{1}{x ln(10)}$
Example 9:

Find the derivative of $y$ = $log_a (x)$

Solution:

$y$ = $log_a (x)$

$y$ = $\frac{ln(x)}{ln(a)}$

$\frac{dy}{dx}$ = $\frac{1}{x ln(a)}$

## Examples

Example 1:

Find the derivative of $sin^{-1} (x^2 - 2x +3)$

Solution:

Let $y$ = $sin^{-1}(x^2 - 2x +3)$

Differentiating with respect to x using chain rule we get

$\frac{dy}{dx}$ = $\frac{1}{\sqrt{1-(x^2 - 2x +3}^2}$ $\frac{d}{dx}$ $(x^2 - 2x +3)$

$\frac{dy}{dx}$ = $\frac{1}{\sqrt{1 - (x^2 -2x +3)^2}}$ (2x-2)

$\frac{dy}{dx}$ = $\frac{2x-2}{\sqrt{1-(x^2 - 2x +3)^2}}$
Example 2:

Find the derivative of $y$ = $tan^{-1}(ln(x))$

Solution:

$y$ = $tan^{-1}(ln(x))$

$\frac{dy}{dx}$ = $\frac{1}{1+(ln(x))^2}$ $\times$ $\frac{d}{dx}$$(ln(x))$

$\frac{dy}{dx}$ = $\frac{1}{1+(ln(x))^2}$ $\times$ $\frac{1}{x}$

$\frac{dy}{dx}$ = $\frac{1}{x + x(ln(x))^2}$
Example 3:

Find the derivative of $csc^{-1}(2x) + 3sec^{-1}(2x)$

Solution:

$y$ = $csc^{-1}(2x) + 3sec^{-1}(2x)$

Differentiate with respect to $x$

$\frac{dy}{dx}$ = $\frac{-1}{(2x)\sqrt{(2x)^2 - 1}}$ $\frac{d}{dx}$ (2x) + $\frac{+3}{(2x)\sqrt{(2x)^2 - 1}}$ $\frac{d}{dx}$ (2x)

$\frac{dy}{dx}$ = $\frac{-2}{(2x)\sqrt{(2x)^2 - 1}}$ + $\frac{+6}{(2x)\sqrt{(2x)^2 - 1}}$

$\frac{dy}{dx}$ = $\frac{-1}{(x)\sqrt{(2x)^2 - 1}}$ + $\frac{+3}{(x)\sqrt{(2x)^2 - 1}}$

$\frac{dy}{dx}$ = $\frac{2}{(x)\sqrt{(2x)^2 - 1}}$

$\frac{dy}{dx}$ = $\frac{2}{x \sqrt{4x^2 -1}}$
Example 4:

Find the derivative of  $y$ = $log(cos^{-1}(x))$

Solution:

Differentiating with respect to $x$ using chain rule

$\frac{dy}{dx}$ = $\frac{1}{cos^{-1}(x) ln(10)}$ $\times$ $\frac{d}{dx}$ $(cos^{-1}(x))$

$\frac{dy}{dx}$ = $\frac{1}{cos^{-1}(x) ln(10)}$ $\times$ $\frac{1}{\sqrt{1-x^2}}$

$\frac{dy}{dx}$ = $\frac{1}{cos^{-1}(x) ln(10)\sqrt{1-x^2}}$