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How can I differentiate $\sin\left(\dfrac{1}{x}\right)$. Do I take $\sin^{-1}(x)$ or what? If I let $u = \dfrac{1}{x}$, then $\sin(u)'$ equals $\cos(u)$, then replace $u = \dfrac{1}{x}$, $\cos\left(\dfrac{1}{x}\right)$.

Martin Sleziak
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hello
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    Hint: Use the chain rule.. $$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$ – Matthew Cassell Feb 24 '15 at 04:29
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    Thanks solved. [cos(1/x)][-1/x^2] – hello Feb 24 '15 at 04:33
  • If you can differentiate $\sin(3x)$ and then $\sin(x^2)$ then do the same steps with $\sin(1/x)$ – MasB Feb 24 '15 at 04:34
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    Do not delete a question once you receive an answer. That deprives the people who helped you a chance for the upvote appreciation. Most likely your action was an honest mistake, but be advised that users react very strongly to that. The reason is that it is easy to imagine very bad motives for self-deletion. – Jyrki Lahtonen Feb 24 '15 at 07:19
  • To expand on what @JyrkiLahtonen said, it is not only because self-deletion (after an answer has been given) deprives answerers of possible upvote appreciation, but it prevents other MSE users in the future from benefiting from your post. It also may be very frustrating for people who answer when their effort to typeset an answer proves fruitless. For example, consider this question with numerous lengthy answers. If OP deleted the question, many people would be upset. Just a heads up. :) – Daniel W. Farlow Feb 24 '15 at 22:16
  • Can anyone give a hint why this very straightforward question got over $20,000$ views? Is there some way a bot is responsible? – David G. Stork May 26 '19 at 21:14

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As Mattos pointed out in a comment, you really want to use the chain rule here; that is, recall that $(f(g(x))'=f'(g(x))g'(x)$.

In your case, your $f(x)$ is $\sin(x)$ and your $g(x)$ is $\frac{1}{x}$; thus, $f(g(x))=\sin(1/x)$. To find $(f(g(x))'$, note that $f'=\cos(x)$ and $g'=-\frac{1}{x^2}$ [I assume you know how to get that far at least]. Thus $$ (f(g(x))'=f'(g(x))g'(x)=\cos(1/x)\cdot(-1/x^2)=-\frac{\cos(1/x)}{x^2}. $$

Daniel W. Farlow
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