Can a derivative of a real valued function have uncountable points of discontinuity?

Question

Suppose $f$ be a real-valued function, such that $f'$ exists everywhere in the domain.
I am thinking about the problem in following steps-
1) Can $f'$ have jump discontinuity?-No, since if it has jump discontinuity, then $f'$ will violate IVP(Darboux's theorem)
2) Can $f'$ have countable number of discontinuity?
3) Can $f'$ have uncountable number of discontinuity?
So, first of all notice that (by ($1$)) $f'$ can have only infinite discontinuity.
I am trying to construct a sequence of function- $f_n:[0,1]\to\Bbb{R}$ by
$f_1(x)= \begin{cases} x^2\sin\left(\frac{\pi}{x+1-{1\over3}}\right), & \text{$x\in[0,{1\over3}]$}\\ (x-{1\over3})^2\sin\left(\frac{\pi}{x+1-{2\over3}}\right), & \text{$x\in[{1\over3},{2\over3}]$}\\ (x-{2\over3})^2\sin\left(\frac{\pi}{x+1-1}\right), & \text{$x\in[{2\over3},1]$}\\ \end{cases} $
My basic motivation is to divide $[0,1]$ into $3^n$ subintervals of equal length for each $n\in\Bbb{N}$. And then define each $f_n$ in the above manner i.e. basically if I want to define $f_n$, then I will divide $[0,1]$ into $3^n$ number of subintervals of equal length, then suppose $[a_n,b_n]$ is one of those sub-intervals. Then define $f_n$ on that perticular $[a_n,b_n]$ by-
$f_n(x)=(x-a_n)^2\sin\left(\frac{\pi}{x+1-{b_n}}\right)$.
Now, I take the whole sequence $\{f_n\}$.
My question is- Is it covergent? If yes and coverge to $f$, then I will take the function $f$. Will $f$ work as the required function? I have no idea about these questions. Even I can't go further. May be my method is wrong.
Can anybody solve this problem? Thanks for assistance in advance.

Answer 1

Hint: As you noticed, the function $x^2\sin(1/x)$ naturally comes to mind here. It may help to do this: Let $C$ denote the Cantor set. Define

$$f(x)= \begin{cases}d(x,C)^2\sin(1/d(x,C)),&x\notin C\\0,&x\in C\end{cases}$$

This $f$ may not quite do it, but it seems close to what you seek.