# Derivative of Inverse Function

Examples with detailed solutions on how to find the derivative (Differentiation) of an inverse function are presented. Also more exercises with answers are included.

## Derivative of Inverse Function Formula (theorem)

Let \( f \) be a function and \( f^{-1} \) its inverse. One of the properties of the inverse function is that \[ f(f^{-1}(x)) = x \] Let \( y = f^{-1}(x) \) so that. \[ f(y) = x \] Differentiate both sides using chain rule on the left side. \[ \dfrac{df}{dy} \dfrac{dy}{dx} = 1 \] Solve for \( \dfrac{dy}{dx} \) \[ \boxed { \dfrac{dy}{dx} = \dfrac{df^{-1}}{dx} = \dfrac{1}{ \dfrac{df}{dy}} } \] which may also be written as \[ \boxed { \dfrac{df^{-1}}{dx} = \dfrac{1}{ f'(f^{-1}(x)) } } \] where \( f' \) is the first derivative of \( f \).

### Example 1

Find the derivative of the inverse of function \( f \) given by \[ f(x)= \dfrac{x}{2} - 1 \]### Solution to Example 1

We present__two methods__to answer the above question. In the first method we calculate the inverse function and then its derivative. In the second method, we use the formula developed above.

#### Method 1

The first method consists in finding the inverse of function \( f \) and differentiate it. To find the inverse of \( f \) we first write it as an equation \[ y = \dfrac{x}{2} - 1 \] Solve for \( x \). \[ x = 2 y + 2 \] Interchange \( x \) and \( y \) to obtain the inverse. \[ y = f^{-1} (x) = 2 x + 2 \] The above gives the inverse function of \( f \) whose derivative is given by \[ \dfrac{dy}{dx} = \dfrac {df^{-1}}{dx} = 2 \]#### Method 2

The second method starts with one of the most important properties of inverse functions.Given \[ f(x)= \dfrac{x}{2} - 1 \] hence \[ f'(x) = \dfrac{1}{2} \]

Substitute \( f' \) by \( \dfrac{1}{2} \) in the formula \( \dfrac{df^{-1}}{dx} = \dfrac{1}{ f'(f^{-1}(x)) } \) to obtain \[ \dfrac{df^{-1}}{dx} = \dfrac{1}{\frac{1}{2}} = 2 \] Note that The first method can be used only if we can find the inverse function explicitly.

### Example 2

Find the derivative \( \dfrac{dy}{dx} \) where \( y = \arcsin x \).### Solution to Example 2

\( \arcsin x \) is the inverse function of \( \sin x \) and hence\[ \sin(\arcsin(x)) = x \qquad (I) \]

Given \[ y = \arcsin x \]

Take the sine of both sides in the above

\[ \sin y = \sin(\arcsin x ) \] Simplify using (I) \[ \sin y = x \] Differentiate both sides of the above equation, with respect to \( x \) , using the chain rule on the left side.

\[ \dfrac {dy}{dx} \cos y = 1 \] Solve for \( \dfrac {dy}{dx} \)

\[ \dfrac {dy}{dx} = \dfrac{1}{\cos y} \] Make the substitution \( y = \arcsin x \) in the above \[ \dfrac {dy}{dx} = \dfrac{1}{\cos (\arcsin x) } \qquad (II) \] Simplify \( \cos \arcsin x \) as using the the trigonometric identity \( \cos x = \sqrt {1 - \sin^2 x} \) by writing \[ \cos (\arcsin x) = \sqrt {1 - \sin^2( \arcsin x)} \] Simplify using the property of inverse fiunctions: \( \sin (\arcsin x) = x \) which gives \[ \cos (\arcsin x) = \sqrt {1 - x^2} \] Substitute in (II) above to obtain the final answer \[ \boxed { \dfrac {dy}{dx} = \dfrac{d(\arcsin(x)}{dx} = \dfrac{1}{ \sqrt {1 - x^2} } } \] Note that the above result could have been obtained using the formula (theorem) above but here we have shown how to find the derivative of the inverse without ( remembering ) the formula.

## Exercises

Find the derivative of the inverse of each function given below.

- \( f(x) = 3x - 4 \)

- \( g(x) = \arccos x \)

- \( h(x) = \arctan x \)

## Answers to the Above Exercises

- \( (f^{-1})'(x) = \dfrac{1}{3} \)

- \( (g^{-1})'(x) = -\dfrac{1}{\sqrt {1-x^2}} \)

- \( (h^{-1})'(x) = \dfrac{1}{\sqrt {1+x^2}} \)

## More References and links