TITLE: How do I differentiate $\sin\left(\frac1x\right)$
QUESTION [5 upvotes]: 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 $u = \dfrac{1}{x}$, $\cos\left(\dfrac{1}{x}\right)$.

REPLY [10 votes]: 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
$$
(f(g(x))'=f'(g(x))g'(x)=\cos(1/x)\cdot(-1/x^2)=-\frac{\cos(1/x)}{x^2}.
$$