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Suppose $E\subseteq\mathbb R^n$ and $f$ maps $E$ into $\mathbb R^m$. Let $g$ map a subset of $\mathbb R^m$ into $\mathbb R^p$. If $f$ is differentiable at $x\in E$ and $g$ is differentiable at $f(x) \in f(E)$, then the composition $g \circ f$ is differentiable at $x$ and $$(g\circ f)'(x) = g'(f(x)) f'(x).$$

Proof. By definition of differentiability, $x$ is an interior point of $E$ and $f(x)$ is an interior point of the domain of $g$. Therefore, continuity of $f$ at $x$ ensures that $x$ is an interior point of the domain of $g \circ f$.

This is the first part of proving that $x$ is an interior point of the domain of the composition function.

my question is why they use the continuity of $f$ at $x$? Where it is trivial that $x$ is in the domain of $f$ which is $E$ and by the hypothesis $x$ is an interior point of $E$ and the domain of $g \circ f$ is also $E$.

This proof is from SHIRALI-BASUDEVA MULTIVARIABLE ANALYSIS BOOK.

Sorry I am not so much accustomed with writing in LaTeX. Please make some edit for me.

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  • $\begingroup$ They meant to write "By our definition of differentiability, $x$ must is an interior point of $E$..." It is usually assumed that we only compute derivatives at interior points, although this is of course not necessary. $\endgroup$ – John B Oct 11 at 16:08
  • $\begingroup$ Yeah but why they use the phrase "continuity of f at X ensure that X is interior point of the domain of g o f " ? $\endgroup$ – LAMDA Oct 11 at 16:12
  • $\begingroup$ It is clear to you that for differentiability of $g\circ f$ at $x$ they have to prove first that $x$ is an interior point of the domain of $g\circ f$? $\endgroup$ – amsmath Oct 11 at 17:15
  • $\begingroup$ @amsmath yeah it is very clear to me that at first I have to prove that $x$ is an interior point of $g\circ f$.. My doubt is in the method to prove this $\endgroup$ – LAMDA Oct 11 at 17:19
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Let $F$ be the domain of $g$. Then $x$ is an interior point of $E$ and $f(x)$ is an interior point of $F$. That is, there exist open sets $U\subset E$ and $W\subset F$ with $x\in U$ and $f(x)\in W$.

Consider $V := f^{-1}(W)$. Then, by continuity of $f$, $V$ is open and so is $U\cap V$. We have $x\in U\cap V$ and $U\cap V\subset E\cap f^{-1}(F)$, which is the domain of $g\circ f$.

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  • $\begingroup$ Is the domain of $g\circ f$ is $E \cap f^{-1}(F)$ ?I thought that it is only $E$.. $\endgroup$ – LAMDA Oct 11 at 17:30
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    $\begingroup$ Of course not. It is $\{x\in E : f(x)\in F\} = E\cap f^{-1}(F)$. $\endgroup$ – amsmath Oct 11 at 17:33

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