Notes on Physics and Math

Calculus

• Derivitive of Inverse Functions

This note explains how to take the derivative of an inverse function.

Statement:

Take a function $f$, and let $g = f^{-1}$. Then we can compute its derivative using:

$\begin{gather*} f(g(x)) = x\\ \implies \dv{}{x}f(g(x)) = 1\\ \implies f'(g(x))g'(x) = 1. \end{gather*}$

So that, remembering $g=f^{-1}$, we get:

$$$\label{eq:inverse} \boxed{g'(x) = \frac1{f\left(f^{-1}(x)\right)}.}$$$

An Example:

You can use Eq. ($\ref{eq:inverse}$) this to work out the derivatives of the inverse trigonometric functions. As an example: \begin{align} \dv{}{x}\arccos x &= \frac1{-\sin(\arccos(x))}, \nonumber\\ &= \frac{-1}{\sqrt{1-x^2}}. \end{align}