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!
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$\begingroup$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 $$y'=-\frac{\sqrt{\cos^2\left(\frac{1}{x}\right)}}{x^2} \\ =-\frac{\cos\left(\frac{1}{x}\right)}{x^2} $$
$\endgroup$ $\begingroup$I dont think anyone would use the limit definition for this. Its more reasonable to use the chain rule.
We know that $d/dx \sin(x) = \cos(x)$ and that $d/dx\; 1/x = -1/x^2$. Together, $$d/dx \sin(1/x) = \sin'(1/x) (1/x)' = \cos(1/x) (-1/x^2)$$
$\endgroup$ $\begingroup$\begin{align*} \lim_{h\rightarrow 0}\frac{\sin(1/(x+h))-\sin(1/x)}{h}&=\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}&=\frac{1}{x^2}\\ \lim_{h\rightarrow 0}\frac{\cos(h/x(x+h))-1}{h}&=0 \end{align*} (Use the substitution $\alpha=h/x(x+h),$ noting that $\alpha\rightarrow 0$ as $h\rightarrow 0$ ).
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