Derivatives of Inverse Functions (Except Trigonometric)

Introduction

An inverse function “undoes” the work of a function.
If $$f(x)$$ maps an input x to an output y, then its inverse function $$f^{-1}(x)$$, maps y back to $x$.

Algebraic properties :

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

Derivative of an Inverse Function — The Rule

If $$f(x)$$ is differentiable and has an inverse function $$f^{-1}(x)$$,
then the derivative of the inverse is given by:

$$(f^{-1})^\prime(x) = \frac{1}{f^\prime(f^{-1}(x))}$$​​

Understanding the Formula

This formula means:

• To find the derivative of the inverse function at a point x, you first find the value of $$f^{-1}(x)$$,
  then plug that into $$f^\prime(x)$$, and finally take its reciprocal.

Derivation of the Formula

Start from the identity:

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

Differentiate both sides with respect to x:

$$f^\prime(f^{-1}(x)) \cdot (f^{-1})^\prime(x) = 1 $$

Solve for $$(f^{-1})^\prime(x)$$:

$$(f^{-1}(x)) = \frac{1}{f^\prime(f^{-1}(x))}$$​

Example 1:

$$f(x) = e^{x}$$

The inverse is: $$f^{-1}(x) = ln(x)$$

Then: $$f^\prime(x) = e^{x}$$

By the inverse formula:

$$(f^{-1}(x)) = \frac{1}{f^\prime(f^{-1}(x))} = \frac{1}{e^{ln(x)}} = \frac{1}{x}$$​​

Example 2:

$$f(x) = x^{3}$$

The inverse is : $$f^{-1}(x) = \sqrt[3]{x}$$​

Then:

$$f^\prime(x) = 3x^{2}$$

Using the inverse rule:

​$$(f^{-1}(x)) = \frac{1}{f^\prime(f^{-1}(x))} = \frac{1}{3(\sqrt[3]{x})^{2}} = \frac{1}{3x^{\frac{2}{3}}}$$​​