# About the chain rule in multivariate calculas

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.

• 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. – John B Oct 11 at 16:08
• 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 " ? – LAMDA Oct 11 at 16:12
• 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 $g\circ f$? – amsmath Oct 11 at 17:15
• @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 – LAMDA Oct 11 at 17:19

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