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I am little confused as to what rule I use to find the derivative of $y=x\sin(1/x)$. Is it the Chain rule or a combination of the product rule and the chain rule? The answer is $\sin (2/x) - (2/x)\cos(2/x)$.

I do not have a clue how they got that answer. Can someone explain with steps?

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    $\begingroup$ Is the problem written correctly? $\endgroup$ – Amzoti Mar 15 '15 at 12:16
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    $\begingroup$ The answer you give would be correct if you replace every $2$ with $1$. $\endgroup$ – Namaste Mar 15 '15 at 12:29
  • $\begingroup$ As you see in answers use chain rule and answer is $\sin(\frac{1}{x})-\frac{1}{x}\cos(\frac{1}{x})$. Answer given by you is certainly wrong, you can verify by putting $x=\frac{2}{\pi}$ $\endgroup$ – Harish Mar 15 '15 at 12:40
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You can use the product rule $f(x)\cdot g(x) = (x)\cdot(\sin(1/x))$

And in finding the derivative of $\sin(1/x)$, you'll need the chain rule. To this end, write $\sin (1/x)$ as $\sin (x^{-1}).$

$$y=x \sin(x^{-1})$$

$$y' = \sin(x^{-1}) + x\cos(x^{-1})(-x^{-2}) = \sin(1/x) + x\cos(1/x)\left(\frac{-1}{x^2}\right)$$

$$y' = \sin(1/x) - (1/x)\cos(1/x) = \sin(1/x) - \frac{\cos(1/x)}{x}$$

Or, you can write as a fraction with common denominator: $$y' = \frac{x\sin(1/x) - \cos(1/x)}{x}$$

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You indeed need to use a combination of the Product rule and the chain rule.

First, apply the product rule:- $$\frac{dy}{dx}=\frac{d}{dx}(x)\sin(1/x)+x\frac{d}{dx}(\sin(1/x))=\sin(1/x)+x\color{blue}{\frac{d}{dx}(\sin(1/x))}$$ Next apply the chain rule to the parts highlighted in blue as follows:- $$\frac{d}{dx}(\sin(1/x))=\frac{d}{d(1/x)}\sin(1/x)\times\frac{d}{dx}(1/x)=\cos(1/x)\times-\frac{1}{x^2}$$ Can you take it from here?

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Hint:

product rule: $$ y'= \sin \left(\dfrac{1}{x}\right)\dfrac{d}{dx}(x)+x\dfrac{d}{dx}\sin \left(\dfrac{1}{x}\right) $$

Chain rule:

$$ \dfrac{d}{dx}\sin \left(\dfrac{1}{x}\right)=\dfrac{d}{dx}\sin \left(x^{-1}\right)= \cos \left(x^{-1}\right) \dfrac{d}{dx} \left(x^{-1}\right) $$

and the answer in OP is wrong.

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