# How do I differentiate $\sin\left(\frac1x\right)$

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

• Hint: Use the chain rule.. $$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$ – Mattos Feb 24 '15 at 4:29
• Thanks solved. [cos(1/x)][-1/x^2] – hello Feb 24 '15 at 4:33
• If you can differentiate $\sin(3x)$ and then $\sin(x^2)$ then do the same steps with $\sin(1/x)$ – Bernard Massé Feb 24 '15 at 4:34
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## 1 Answer

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