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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?

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  • $\begingroup$ 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. $\endgroup$ – Alex R. Mar 25 '19 at 18:25
  • $\begingroup$ Oh, I got that wrong, thanks for the counterexample. Can't see what should be the necessary condition then. $\endgroup$ – Leonardo V. Sailer Mar 25 '19 at 18:40
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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.

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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 $$\lim_{h\to 0}g(x_0)=g(x_0)$$

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