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 $u = \dfrac{1}{x}$, $\cos\left(\dfrac{1}{x}\right)$.

  • 11
    $\begingroup$ Hint: Use the chain rule.. $$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$ $\endgroup$
    – mattos
    Feb 24, 2015 at 4:29
  • 1
    $\begingroup$ Thanks solved. [cos(1/x)][-1/x^2] $\endgroup$
    – hello
    Feb 24, 2015 at 4:33
  • $\begingroup$ If you can differentiate $\sin(3x)$ and then $\sin(x^2)$ then do the same steps with $\sin(1/x)$ $\endgroup$ Feb 24, 2015 at 4:34
  • 11
    $\begingroup$ 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. $\endgroup$ Feb 24, 2015 at 7:19
  • $\begingroup$ 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. :) $\endgroup$ Feb 24, 2015 at 22:16

1 Answer 1


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 $$ (f(g(x))'=f'(g(x))g'(x)=\cos(1/x)\cdot(-1/x^2)=-\frac{\cos(1/x)}{x^2}. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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