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

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.

• – José Carlos Santos Mar 28 '19 at 19:41
• Either 2) or 3) need to be edited. – user Mar 28 '19 at 20:01

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.
• What do you mean by $d$ here? Distance? – Biswarup Saha Mar 28 '19 at 20:19
• Yes, $d(x,C)$ is the distance from $x$ to $C.$ – zhw. Mar 28 '19 at 20:22
• You are saying this $f$ may not work, but close to the answer. Basically, I can't generalize(even visualize) it. Can you generalize the function explicitly so that it will work as the required answer? your answer is likely to be work, but I can't prove it – Biswarup Saha Mar 28 '19 at 20:33
• It's a bit of a challenging problem. You have to expect to spend some time on it. Can you show $f'(x)=0$ for every $x\in C?$ – zhw. Mar 28 '19 at 20:50