Notes on Physics and Math


Condensed Matter Physics

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:

    \[\begin{equation} \label{eq:inverse} \boxed{g'(x) = \frac1{f\left(f^{-1}(x)\right)}.} \end{equation}\]

    An Example:

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

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