When differentiability of the product implies differentiability of the individual terms?

Say we have $ h(x)=f(x)\cdot g(x)$ where $f$ and $g$ are continuous and strictly increasing. It follows they are differentiable almost everywhere and so is $h$. We also know that $f>0$ and $g>0$. I'm trying to find a straightforward proof that under these conditions, if $h$ is differentiable everywhere then both $f$ and $g$ are also differentiable everywhere. I have more structure on these functions but I was hoping I did need to impose additional assumptions.

Let $f(x)=g(x)=e^x$ for $x \le 0$, while for $x>0$ let $f(x)=1+(1/2)x$ and $g(x)=1+(3/2)x$. Then $h(x)=e^{2x}$ when $x \le 0$ and for $x>0$ it's $h(x)=1+2x+3x^2/4.$

So in this case neither of $f,g$ are differentiable at $0$, while $h$ is differentiable everywhere. (These functions are each positive and strictly increasing everywhere.)

Here is an ugly example:

Take $h(x) = (x+1)^6+1$, $g(x) = 1+2x+|x|$ on $|x| \le {1 \over 2}$.

Now let $f(x) = {h(x) \over g(x)}$.

All are strictly increasing, strictly positive and continuous. $g$ (and hence $f$) is not differentiable at $x=0$.