# Doubt in The inverse function theorem of Rudin's Principles of mathematical Analysis (Theorem 9.24)

Theorem 9.24: Suppose $\textbf{f}$ is a $C'-$ mapping of an open set $E \subset \mathbb{R^n}$ into $\mathbb{R^n}$.$\textbf{f} \thinspace^\prime\textbf{ (a)}$ is invertible for some $\textbf{a}\in E$ and $\textbf{b=f(a)}$ then

(a) there exist open sets $U$ and $V$ in $\mathbb{R^n }$ such that $\textbf{a}\in U, \textbf{b}\in V,\textbf{f}$ is one-to one on $U$ and $\textbf{f}(U)=V$

(b) if $\textbf{g}$ is the inverse of $\textbf{f}$[which exist , by (a)], definedin $V$ by $\textbf{g(f(x))=x }$$\space$ $(\textbf{x}\in U)$ then $\textbf{g}\in C'(V)$

To prove (a) Rudin took a function $\varphi$ which is defined as to each $\textbf{y}\in \mathbb{R^n },\varphi(\textbf{x})=\textbf{x}+A^{-1}(\textbf{y}\thinspace \textbf{-f(x))}\space \space (\textbf{x}\in E )$(Equation 48) Then he states that $\varphi '(\textbf{x})=I- A^{-1}\textbf{f}\space'(x)$.I didn't got how rudin found $\varphi '(\textbf{x})$.Please help me to understand.

• That looks like a basic calculus computation. What is the source of your doubt? Jul 4, 2018 at 17:21
• Sorry I am not understanding this by basic calculus computation.Please help me Jul 4, 2018 at 17:29

Note that $$\varphi(x)$$ is the sum of $$x$$ with $$A^{-1}\bigl(y-f(x)\bigr)$$. Well, $$x$$ is just the identity function, and $$A^{-1}$$ is linear. Finally, $$y$$ is constant. Therefore, by the rule for differentiating sums and by the chain rule (together with that fact that $$(A^{-1})'=A^{-1}$$), we get that\begin{align}\varphi'(x)&=\operatorname{Id}+A^{-1}\circ(y-f)'(x)\\&=\operatorname{Id}+A^{-1}\bigl(-f'(x)\bigr)\\&=\operatorname{Id}-A^{-1}\bigl(f(x)\bigr).\end{align}