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.
Best Answer
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}