If the composition of two functions is differentiable, and one of them is differentiable, must the second one also be differentiable?

Question

Suppose that we have three distinct functions $f,g,h:\mathbb{R}\rightarrow\mathbb{R}$, that $f$ and $g$ are differentiable, and that $f=g \circ h$, where $\circ$ denotes composition.

Does it necessarily follow that $h$ must be differentiable too?

Answer 1

This is a modification of the hint about Constant function , given earlier by user MPW , though I will give Non-Constant Example.

Let $f(x)=g(h(x))$ where $g(x)=x^2$ & $h(x)$ is like this :

$h(x) = +x \text{ [ if x is rational ] }$
$h(x) = -x \text{ [ if x is irrational ] }$

In general , we can check that $h$ has no Derivative , while $g$ has Derivative $2x$
More-over , $f(x) \equiv (\pm x)^2 \equiv x^2$ , thus $f$ will have Derivative $2x$

This is a Counter-Example to the Claim.

HIGHLIGHT :
Even when $h$ is erratic/chaotic/weird , having no Derivative , "smooth" $g$ can "eliminate" that erraticness/chaos/weirdness to give "smooth" $f$.

Answer 2

Hint:

Consider $\;f(x):=|\cos x|\;$ at $\;x=0\;$ .