Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $x \ne 0$.

share|improve this question
1  
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
2  
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

2 Answers 2

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.

share|improve this answer

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.