# On differentiability (product of two functions)

Given that the product $$f(x)\cdot g(x)$$ is differentiable at $$x_0$$, I need to work out what conditions over $$f(x)$$ guarantee that $$g(x)$$ is also differentiable at $$x_0$$. Note: both $$f$$ and $$g$$ are $$\mathbb{R} \rightarrow \mathbb{R}$$.

My thoughts so far: Well I can easily come up with conditions that guarantee $$g(x)$$ is differentiable on $$x_0$$, for example, if $$f(x)$$ is constant then $$(f\cdot g)'$$ will only exist if $$g'$$ exists. But that's probably not a necessary condition.

I wrote $$(f\cdot g)'$$ as $$\lim_{h\to 0} \left[f(x_0+h)\cdot\frac{(g(x_0+h)-g(x_0))}{h} +g(x_0)\cdot\frac{(f(x_0+h)-f(x_0))}{h}\right]$$ and that gives me a hint that maybe $$f(x)$$ being differentiable at $$x_0$$ is sufficient and necessary, but i can't prove it, could anyone please help me out?

• Differentiability of $f$ is not enough. For example, take $f(x)=x^2$ and $g(x)=x$ for $x<0$ and $g(x)=2x$ for $x\geq 0$, and work out the derivative of $f\cdot g$ at 0 using left/right derivatives. – Alex R. Mar 25 '19 at 18:25
• Oh, I got that wrong, thanks for the counterexample. Can't see what should be the necessary condition then. – Leonardo V. Sailer Mar 25 '19 at 18:40

Maybe I'm not seeing the trees for the forest, but to me a 'natural' condition would be that $$f$$ is differentiable at $$x_0$$ and $$f(x_0) \neq 0$$. That would make $$g$$ the quotient of (differentiable at $$x_0$$) functions $$fg$$ and $$f$$, and thus the quotient rule would apply.
The differentiability part for $$f$$ at $$x_0$$ is necessary if $$g(x_0) \neq 0$$, with a similar reason.
As has been remarked in the comments, if $$f(x_0)=0$$, then $$g$$ may or may not be differentiable (at $$x_0$$) even if $$f$$ is.
You must use that $$\lim_{h\to 0}\frac{g(x_0+h)-g(x_0)}{h}=g'(x_0)$$ and $$\lim_{h\to 0}\frac{f(x_0+h)-f(x_0)}{h}=f'(x_0)$$ And $$\lim_{h\to 0}f(x_0+h)=f(x_0)$$ and $$\lim_{h\to 0}g(x_0)=g(x_0)$$
• Given is that $$(fg)$$ is differentiable in $$x_0$$ – Dr. Sonnhard Graubner Mar 25 '19 at 18:45