contradiction to chain rule.

Question

As the chain rule states:

If $f(u)$ is differentiable at point $u=g(x)$, and $g(x)$ is differentiable at $x$, then the composite function $(f\circ g)(x)=f(g(x))$ is differentiable at $x$ , and $$(f\circ g)'(x)=f'(g(x)).g'(x).$$

Then the question says:

We are given: $f(x)=x^2$ and $g(x)=|x|$. Then the composites $(f\circ g)(x)=|x|^2=x^2$ and $(g\circ f)(x)=|x^2|=x^2$ are both differentiable at $x=0$ even though $g$ itself is not differentiable at $x=0$. Does this contradict the chain rule?

I think it's not so because chain rule says that $f(u)$ is differentiable at $u=g(x)$, and $g(x)$ is differentiable at $x$, and does not imply vice-versa. Am I correct?

Answer 1

Yes. It does not contradict the chain rule.

The chain rule states $P\cap Q \implies R$, it does not mean that $R\implies P\cap Q$.

$P$ and $Q$ are sufficient conditions for $R$, they are not necessary conditions.

Where $P$ is "$f$ is differentiable at $g(x)$", $Q$ is "$g$ is differentiable at $x$", and $R$ is "$f\circ g$ is differentiable at $x$".