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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.

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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 $D\psi_q=[Df_p]^{-1}$.

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  • $\begingroup$ Thank you! That clarifies a lot — I still get confused with differential notation pretty often. One question: why is the derivative of $d\mathbf{f}^{-1}_{\mathbf{p}}(\mathbf{y}-\mathbf{q}) = d\mathbf{f}^{-1}_{\mathbf{p}}$? Where does the $(\mathbf{y}-\mathbf{q})$ part actually "go"? $\endgroup$ Commented Feb 18, 2023 at 21:26
  • $\begingroup$ @AlexanderPope There’s two ways of looking at it. First, $[Df_p]^{-1}(y-q)=[Df_p]^{-1}(y)-[Df_p]^{-1}(q)$ due to linearity. The second term is a constant, so when differentiating it becomes zero. Hence, we only have $Df_p^{-1}$. $\endgroup$
    – peek-a-boo
    Commented Feb 18, 2023 at 21:36
  • $\begingroup$ The second way of viewing things is via the chain rule. Write $g(y)=y-q=I(y)-q$, where $I$ is the identity map. So, $Dg_y=DI_y-0=I$ (the zero is because we’re differentiating a constant $q$, and $DI_y=I$ because $I$ is a linear map so at every point it is it own derivative). Now, $\psi(y)=\left([Df_p]^{-1}\circ g\right)(y)$. So, by the chain rule, $D\psi_y=D\bigg([Df_p]^{-1}\bigg)_{g(y)}\circ Dg_y=\bigg([Df_p]^{-1}\bigg)\circ I=[Df_p]^{-1}$ based on what we said. $\endgroup$
    – peek-a-boo
    Commented Feb 18, 2023 at 21:40
  • $\begingroup$ So, the $(y-q)$ part actually goes to the point of evaluation of the derivative of $[Df_p]^{-1}$, but since we’re differentiating a linear map, ‘the point of evaluation is not important. But, if for example you had a more complicated function like $\psi(y)=\phi(y-q)$, where $\phi$ is merely differentiable (not necessarily linear), then by the chain rule and “sum rule”, we have $D\psi_y=D\phi_{y-q}\circ (DI_y-0)=D\phi_{y-q}$. So, that’s where the $(y-q)$ “goes”. But for now, I think my first comment will provide you the most direct answer. $\endgroup$
    – peek-a-boo
    Commented Feb 18, 2023 at 21:42
  • $\begingroup$ Thank you again, @peek-a-boo. So in other words the derivative of $d\mathbf{f}^{-1}_{\mathbf{p}}(\mathbf{y}-\mathbf{q})$ is actually $ d\mathbf{f}^{-1}_{\mathbf{p}}(\mathbf{y})$ However, we omit the $\mathbf{y}$ as the point of evaluation is irrelevant? $\endgroup$ Commented Feb 18, 2023 at 21:57

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