Derivative of Inverse Differential in Proof of Inverse Function Theorem.

Question

On p. $194$ of Mathematical Analysis by Andrew Browder he defines a map $\psi: \mathbb{R}^n \rightarrow \mathbb{R}^n$ by $$\psi(\mathbf{y})=d\mathbf{f}^{-1}_{\mathbf{p}}(\mathbf{y}-\mathbf{q})$$ as a tool to prove the inverse function theorem. He then claims that $$\psi'(\mathbf{q})=[\mathbf{f}'(\mathbf{p})]^{-1},$$ where $\mathbf{f}(\mathbf{p})=\mathbf{q}$. Why is this the case? Isn't $d\mathbf{f}^{-1}_{\mathbf{p}}$ a linear map, and therefore its derivative should just be itself? But where is the $\mathbf{q}$ on the right hand side of the equation? Many thanks.

Answer 1

Indeed, $\psi$ is an affine transformation (linear plus a constant), so at every point $y$, the derivative $D\psi_y$ is equal to the linear part of $\psi$, which in this case is just $(Df_p)^{-1}$. Since $D\psi_y=[Df_p]^{-1}$ is true for all $y$, it is in particular true when $y=q$, thus $D\psi_q=[Df_p]^{-1}$.