# If $f(x) = h(x)g(x)$, is $h$ differentiable if $f$ and $g$ are?

I know that if I have two differentiable functions $f, g$ then the functions $(f + g)$ and $fg$ are also differentiable. I would like to find a way how to argue about the function $h$ where \begin{equation} f(x) = (hg)(x) := h(x)g(x) \quad \text{and } f,g \text{ are differentiable} \end{equation}

For a start I can conclude $h$ is differentiable at all points where $g(x) \neq 0$ since there I can express $h$ as \begin{equation} h = \frac{f}{g} \end{equation}

But for the remaining points I am not sure, my guess is that $h$ is differentiable, any hints how I can make this into a formal argument ? Or am I probably wrong ? In that case, would it help to impose further smoothness on $f$ and $g$, say both are $C^\infty$ ?

Many thanks!

## 2 Answers

In general the answer is no: Consider $f(x) = x^3$, $h(x) = |x|$ and $g(x) = \text{sign}(x) x^2$. In this case, $f$ is differentiable at $0$ while $h$ is not.

I guess that $C^\infty$ in not enough, but analytic functions should work if $f$ is not constant zero.

• that's a very interesting counterexample, great answer many thanks ! – harlekin Feb 1 '12 at 10:44
• I swapped $f$ and $g$ in your answer to match the OP's notation. Please check that I didn't accidentally introduce any mistakes. Oh, and +1. :) – Ilmari Karonen Feb 1 '12 at 15:42

No. Let $f(x)=g(x)=0$ for all $x$. There are no functions more regular than constant functions. Then $f=gh$ holds for any function $h$, even if it is nowhere-differentiable.

• of course you are right, thanks! – harlekin Feb 1 '12 at 10:44