# 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

- Inverse Function
- differentiation and derivatives