# 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.)

• This $h$ is not $C^\infty$ (let alone real analytical) in $x=0$, of course. Maybe no counter-example exists if it is required that $h$ is analytical? – Jeppe Stig Nielsen Jul 14 '14 at 10:17
• @Jeppe The requirement of OP was only differentiable, and this $h$ is so. I'll try for a smooth $h$ only because it could be considered a "better" example. --Actually I see that copper.hat's example has its $h(x)$ analytical, even a polynomial (with the domain restricted to $|x|\le 1/2$). – coffeemath Jul 14 '14 at 13:51

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 $x=0$.

• Wow. Neat example. How'd you come up with it? – Cameron Williams Jul 14 '14 at 1:30
• Trial & error unfortunately :-). – copper.hat Jul 14 '14 at 2:43
• Yikes! You're a more patient man than I. – Cameron Williams Jul 14 '14 at 2:46
• I doubt it :-). – copper.hat Jul 14 '14 at 3:09
• Why the downvote? – copper.hat Dec 27 '14 at 8:43