Prove the derivative of $\sin(1/x)$ exists How do I prove the derivative of $$\sin(1/x)=-\frac{1}{x^2}\cos(1/x)$$? I understand that you use $$f'(x_0) = \lim_{x \to x_0} \frac{\sin(1/x) - \sin(1/x_0)}{x-x_0} = -\frac{1}{x_0^2}\cos(1/x_0)$$ somehow but am not sure how to show it directly. 

Any guidance would be greatly appreciated!
You could try implicit differentiation to solve this. Let $y = \sin\left(\frac{1}{x} \right)$. Then $\arcsin(y) = \frac{1}{x}$ and $$\frac{dy}{dx}\arcsin(y)=\frac{dy}{dx}\frac{1}{x} \\ \implies \frac{y'}{\sqrt{1-y^2}}=-\frac{1}{x^2} \\ \implies y' = -\frac{\sqrt{1-y^2}}{x^2}$$ Now plug back in $y = \sin\left(\frac{1}{x}\right)$ to get $$y'=-\frac{\sqrt{1-\sin^2\left(\frac{1}{x}\right)}}{x^2}$$ Then use the Pythagorean identity $$1-\sin^2\left(\frac{1}{x}\right)=\cos^2\left(\frac{1}{x}\right)$$ to conclude that $$y'=-\frac{\sqrt{\cos^2\left(\frac{1}{x}\right)}}{x^2} \\ =-\frac{\cos\left(\frac{1}{x}\right)}{x^2}  $$
\begin{align*}
\lim_{h\rightarrow 0}\frac{\sin(1/(x+h))-\sin(1/x)}{h}&amp;=\lim_{h\rightarrow 0}\frac{\sin(1/x-h/x(x+h))-\sin(1/x)}{h}
\end{align*}
Now use $\sin(A-B)=\dots$ to reduce the problem to showing the following:
\begin{align*}
\lim_{h\rightarrow 0}\frac{\sin(h/x(x+h))}{h}&amp;=\frac{1}{x^2}\\
\lim_{h\rightarrow 0}\frac{\cos(h/x(x+h))-1}{h}&amp;=0
\end{align*}
(Use the substitution $\alpha=h/x(x+h),$ noting that $\alpha\rightarrow 0$ as $h\rightarrow 0$ ).