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I am trying to show that this function is classically differentiable. Now I can find the derivative no problem: $2x\sin(\tfrac{1}{x})-\cos(\tfrac{1}{x})$, and this gives me something to aim for, but getting to this using the classical definition of differentiability is proving challenging. Specifically, I am having trouble simplifying the expression $$\tfrac{x^2\sin(\tfrac{1}{x})-a^2\sin(\tfrac{1}{a})}{x-a}$$ I have tried using long division, but I feel like I am going in circles. I've also tried thinking about what I would need to multiply $x-a$ by to get the numerator, but I am stuck there too. Any suggestions?

Bernard
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    For $x\ne 0$, this function is differentiable as a product and composition of differentiable functions. The only case you need to check is at $x=0$. – Yuval Gat Mar 23 '19 at 21:07
  • What is classically differentiable? – Randall Mar 23 '19 at 21:10
  • I am currently doing a research project on the strong (or strict) derivative. I've been using the term "classically differentiable" to differentiate (no pun intended) between when I am talking about a strong derivative or the derivative most of us learn in Calculus. –  Mar 23 '19 at 21:13
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    I don't think this is a duplicate: the O.P. wants to prove differentiability at every point, not only $0$, calculating the limit of the rate of variation when $x\to a$. Wher can we vote to re-open the question? – Bernard Mar 23 '19 at 21:28
  • I was going to say, while what Yuval said is true, it seems to be a way around doing work I may be expected to show in my research paper. I am going to ask my professor what he expects for this particular problem. –  Mar 23 '19 at 21:32
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    For direct evaluation of the limit use $\frac{x^2 \sin \frac{1}{x}- a^2 \sin \frac{1}{a}}{x-a}= \frac{x^2 - a^2}{x-a} \sin \frac{1}{x} + \frac{a^2}{x-a}\left( \sin\frac{1}{x} - \sin\frac{1}{a}\right)$. Then apply the identity $\sin a - \sin b = 2 \sin \frac{a-b}{2} \cos \frac{a+b}{2}$ and ultimately you can evaluate the limit using only continuity and the basic limit $\frac{\sin x}{x} \to 1$ as $x \to 0$. – RRL Mar 23 '19 at 21:44
  • Also, I voted to reopen. – RRL Mar 23 '19 at 21:49
  • Thank you RRL! This is more like what I was looking for and most likely what I will end up needing. –  Mar 23 '19 at 21:52
  • I agree with that, given that the question is asking for beyond differentiability at 0, and have also voted for reopen. – Moya Mar 23 '19 at 21:52
  • Voted to reopen. RRL should pursue the answer. – Randall Mar 23 '19 at 23:44
  • Update: I was able to speak with my professor today, and for the purposes of the question he had asked me, he was only interested in what was happening at zero, like what Yuval was getting at. However, he also showed interest in RRL's response and thought that his answer might be worth putting into my paper as well, though it wouldn't be required... so I do hope that this will be re-opened eventually. –  Mar 25 '19 at 23:33

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