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If I am asked to prove that the function $x^3 \sin(\frac{1}{x})$ is differentiable for all $x \ne 0$, is it sufficient to say something like the following:

We see that $x^3$, $\sin(\frac{1}{x})$ and $\frac{1}{x}$ are differentiable for all $x \ne 0$. It follows, therefore, that their product is also differentiable for all $x \ne 0$.

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Yes, that suffices. Interestingly enough, it is also differentiable at $x=0$. – Pedro Tamaroff Nov 30 '12 at 23:54
@PeterTamaroff Ah, yes, it is. I just made up the function on the spot and I just assumed it wouldn't be. Seeing as this is a complete answer, should you post it as such, I will be happy to accept it. – providence Nov 30 '12 at 23:56
There should be minor wording changes. The sine function is differentiable everywhere, and $1/x$ is differentiable when $x\ne 0$, so their composition (not product) is differentiable at $x\ne 0$. And $x^3$ is differentiable, so the product $\dots$. – André Nicolas Dec 1 '12 at 0:00
up vote 2 down vote accepted

Yes, that suffices. Interestingly enough, it is also differentiable at $x=0$ if one defines $f(0)=0$ to make it continuous.

ADD To add a little more, the rules are:

$(1)$ If $f$ and $g$ are differentiable at $x=a$, so is $f\cdot g$ and $f\dot g)'(a)=f'(a)\cdot g(a)+f(a)\cdot g'(a)$

$(2)$ If $g$ is differentiable at $x=a$ and $f$ is diferentiable at $x=g(a)$ then $f\circ g$ is differentiable at $x=a$ and $(f\circ g)'(a)=f'\circ g(a)\cdot g'(a)$

So you ought to be clear about what you're using to prove that.

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You would also have to say something about the composition of differentiable functions being differentiable, but essentially you're correct.

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