Inspired by Kishore Shenoy

The standard example used is:

$f(x) = \begin{cases} x^2 \sin \frac{1}{x} & x \neq 0 \\ 0 & x = 0 \\ \end{cases}$

It is easy to see that away from $x = 0$ , the function is differentiable, and has derivative of the form $2 x \sin \frac{1}{x} - \cos \frac{1}{x}$ .

At the origin, it is easy to see (esp from the graph as it's bounded by $\pm x^2$ ) that the derivative is 0. We can show this properly using first principles, since
$\lim \frac{ x^2 \sin \frac{1}{x} - 0 } { x } = \lim x \sin \frac{1}{x} = 0.$

We see that the derivative is discontinuous at 0, because it looks like the topologists sine curve.

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This gives a single point of discontinuity. Volterra's function is a fractal version of this construction, and results in a function that is differentiable, but whose derivative is discontinuous on a set of positive measure (due to the fat Cantor set).

I think there is an answer in the link below

Derivative