Initially, I constructed a function with the help of famous Thomae's function, but later I found it to be wrong. Firstly, I did-
Let $f:[0,1]\to\Bbb{R}$ be the Thomae's function defied by $$f(x)=\begin{cases}0& \text{if $x$ is irrational} \\{1 \over q} &\text{if $x$ is rational, $x={p \over q}$ with $p\in\Bbb{Z}$, $q\in\Bbb{N}$ are coprime}\end{cases}$$ This function is Riemann Integrable on $[0,1]$. Now, define $F:[0,1]\to\Bbb{R}$ by $F(x)=\int_{[0,x]} f\:\forall x\in[0,1]$
But here $\int_{[0,x]} f=0\:\forall x\in[0,1]\implies F(x)=0\:\forall x\in[0,1]\implies F$ is differentiable everywhere on $[0,1]$.
But, as per the question $F$ is not the required function.
Can anybody give me an example of such a function with some proper arguments?
Thanks for your assistance in advance!
I provide here an explicit construction (in a more general case).
Consider $D = (a_n)_{n \ge 1}$ a countable subset of $[0,1]$. Then the following function is differentiable exactly on $[0,1] \backslash D$ : $$ f : x \in [0,1] \mapsto \sum\limits_{n=1}^{+\infty} \frac{|x-a_n|}{2^n}.$$
In the rest of the proof, $[a,]$ will denote the points between $a$ and $b$ regardless of which is greater (even if $a > b$, $[a,]$ is not empty ; $[a,b]=[b,a]$). Take any $x \in [0,1]$. Note that if $a > \max(x,y)$ then $|y-a|-|x-a|=x-y$, if $a < \min(x,y)$ then $|y-a|-|x-a|=y-x$, and if $a$ is between $x$ and $y$, $||y-a|-|x-a||\le 2|y-x|$.
Case 1: $x \notin D$. Let $n_0 \in \mathbb{N}$. For $y$ in some neighborhood of $x$, $[x,y]$ does not contain $a_1,...,a_{n_0}$. Then $\frac{f(y)-f(x)}{y-x} = \sum\limits_{a_n > \max(x,y)} \frac{1}{2^n} + \sum\limits_{a_n < \min(x,y)} \frac{-1}{2^n} + \sum\limits_{a_n \in [x,y]} \frac{|y-a_n|-|x-a_n|}{2^n}$. And $\big| \sum\limits_{a_n \in [x,y]} \frac{|y-a_n|-|x-a_n|}{2^n|y-x|}\big| \le \sum\limits_{n>n_0} \frac{2}{2^n}=\frac{1}{2^{n_0-1}}$. Hence $f$ is differentiable at $x$ and $f'(x) = \sum\limits_{a_n > x} \frac{1}{2^n}-\sum\limits_{a_n < x} \frac{1}{2^n}$.
Case 2: $x = a_m \in D$. Let $n_0 \in \mathbb{N}$. For $y$ in some neighborhood of $x$, $]x,y]$ does not contain$a_1,...,a_{n_0}$. Then as before, we have $\frac{f(y)-f(x)}{y-x}=\sum\limits_{a_n > \max(x,y)} \frac{1}{2^n} - \sum\limits_{a_n < \min(x,y)} \frac{1}{2^n} + \frac{|y-a_m|}{2^m (y-a_m)} + C(y)$ where $|C(y)| \le \frac{1}{2^{n_0-1}}$. Denoting $K = \sum\limits_{a_n>x}\frac{1}{2^n}-\sum\limits_{a_n<x}\frac{1}{2^n}$, we find that the left limit is $\frac{f(y)-f(x)}{y-x} \underset{y\to x^-}{\longrightarrow} K - \frac{1}{2^m}$, while the right limit is $\frac{f(y)-f(x)}{y-x} \underset{y \to x^+}{\longrightarrow} K - \frac{1}{2^m}$. Hence $f$ is not differentiable at $x$.
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Conclusion given $D = \{a_n \ |\ n \ge 1\}$ a countable subset of $[0,1]$, $f : x \mapsto \sum\limits_{n=1}^{+\infty}\frac{|x-a_n|}{2^n}$ is differentiable exactly at $[0,1]\backslash D$.