Derivative of the inverse function

Statement

Let $V, W \subseteq \mathbb{R}^n$ be open, and let $f \colon V \to W$ be a bijection. Suppose that $f$ is differentiable at $a \in V$ and that $f^{-1}$ is differentiable at $b = f(a)$. Then $Df(a)$ is invertible, and $$D(f^{-1})(b) = \big( Df(a) \big)^{-1}.$$ In terms of Jacobian matrices, $J_{f^{-1}}(b) = J_f(a)^{-1}$.

Remarks

Proof: apply the chain rule to $f^{-1} \circ f = \mathrm{id}_V$ and $f \circ f^{-1} = \mathrm{id}_W$. This gives $D(f^{-1})(b) \, Df(a) = I$ and $Df(a) \, D(f^{-1})(b) = I$. The hard part is the differentiability of $f^{-1}$. The inverse function theorem gives it when $f$ is of class $C^1$ and $Df(a)$ is invertible. For polar coordinates $f(r, \theta) = (r \cos \theta, r \sin \theta)$ with $r > 0$, $J_f = \begin{pmatrix} \cos \theta & -r \sin \theta \\ \sin \theta & r \cos \theta \end{pmatrix}$, and the Jacobian matrix of a local inverse is the inverse of this matrix.

Explanations

L1 human

Undoing a map undoes its linear approximation. So the derivative of the inverse is the inverse matrix of the derivative, taken at the matching point.