Derivative of the inverse function

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.