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Let $$f(x) = \begin{cases} x^2\sin\frac{1}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases}$$, then at $$x = 0$$
At $$x = 0$$: $$f'(0) = \lim_{h\to 0}\frac{h^2\sin(1/h)}{h} = \lim_{h\to 0} h\sin(1/h) = 0$$. So $$f$$ is differentiable at $$x = 0$$.
For $$x \neq 0$$: $$f'(x) = 2x\sin(1/x) - \cos(1/x)$$. As $$x \to 0$$, $$\cos(1/x)$$ oscillates, so $$\lim_{x\to 0} f'(x)$$ does not exist.
Therefore, $$f$$ is differentiable at $$x = 0$$ but $$f'$$ is not continuous at $$x = 0$$.
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