calculus - Find a $f$ function such that$f'(x)geq 0$ but not continuous

I just started reading continuity, differentiability etc. So I was thinking of an example of following type:

Let $f: [a,b]\rightarrow \mathbb{R}$, where $f(x)$ is monotonically increasing continuous function and differentiable on $(a,b)$ or simply $f'(x)\geq 0$ for $x\in (a,b)$.

can we find such function for which $f'(x)$ is not continuous?

I could not find any, whatever function I take $f'(x)$ is becoming continuous. Is there such function even exists!

Answer

For every $x\in (-1,1).$ Define,

$$f(x) = 100x+ x^2\sin\frac{1}{x} \qquad\text{with $f(0)=0$}$$
which is differentiable and
$$f'(x) = 100+ 2x\sin\frac{1}{x}-\cos\frac{1}{x} \qquad\text{with $f'(0)=100$}$$

$f'$ is not continuous a $x=0$ and you might choose the interval $[a,b]$ around $x=0$ as it suite to you. Indeed

$$f'(x)=100+2x\sin\frac{1}{x}-\cos\frac{1}{x} \ge 100+ 2x\sin\frac{1}{x}-1$$

Since $|\sin a| \le |a|$ and $-1\le-\cos a\le 1$ then $|2x\sin\frac{1}{x}|\le 2$ i.e

$$2x\sin\frac{1}{x}\ge -2$$

therefore

$$f'(x)=100+2x\sin\frac{1}{x}-\cos\frac{1}{x} \ge 100-2-1= 97>0$$ for every $x\in \mathbb R$. so $f$ is increasing and $f'$ is not continuous at $x=0$.