I can't find the derivative of xsin (pi / x) in this question. $$ f(x) = \begin{cases} x \sin \frac{\pi}{x} & x \neq 0\\ 0 & x = 0 \end{cases}. \quad \text{Compute }f'(0). $$
There must also be a differentiability rule for f '(x) to exist.
How do I solve it?
$$f'(x)=\lim_{h \to 0} \frac{h \sin (\pi/h)-0}{h}= \lim_{h \to 0} \sin (\pi/h)=\text{does not exist}.$$
$L=\lim_{x \to 0} \sin(\pi/x)$ does not exist for two sequeces $x_n=\frac{1}{n}$ and $x'_n=\frac{1}{(n+1/2)}$, let $g(x)=\sin(\pi/x)$, then $g(x_n)=0, g(x'_n)=1$ these being different there are two accumulation points. So, the limit $L$ does not exist. Hence, $f(x)$ is not differentiable at $x=0$.