# Inverse Function Rule **A formula for the [[derivative]] of the inverse of a function.** >[!Example] Inverse Function Rule >If $f(x)=y$ and $f^{-1}(y)=x$, then the **inverse function rule** states that >$[f^{-1}]'(y)=\frac{1}{f'(f^{-1}(y))}=\frac{1}{f'(x)}$ The graphs of a function and its inverse are *reflections* in the line $y=x$ and thus their gradients are *multiplicative inverses* of each other. ## Proof A proof of the inverse function rule uses the [[chain rule]]. Let $f$ be an invertible function, where $x$ is in the domain of $f$ and $y$ is in the codomain of $f$ such that $f(x)=y$ and $f^{-1}(y)=x$. The proof can begin with either $f(f^{-1}(y))=y$ or $f(f^{-1}(x))=x$. $f(f^{-1}(y))=y$ Differentiating both sides with respect to $y$ gives $\frac{d}{dy}f(f^{-1}(y))=\frac{d}{dy}y$ Applying the *chain rule* on the left side gives $\begin{align*} \frac{d}{d(f^{-1}(y))}f(f^{-1}(y))\;\frac{d}{dy}f^{-1}(y)&=1 \qquad \qquad \qquad \qquad \enspace \> \\ f'(f^{-1}(y))\;[f^{-1}]'(y)&=1 \\ \therefore [f^{-1}]'(y)&=\frac{1}{f'(f^{-1}(y))} \end{align*}$