$f>0$, $f'\geq 0$ but not continuous. This question is continuation of this question. (And last of this type)

Let $f: [a,b]\rightarrow \mathbb{R}$, where $f(x)$ is monotonically non-decreasing positive 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,obviously. Is there such function even exists!
 A: Take $x\in(-1,1)$
$$ f(x) =100000+ 100x+ x^2\sin\frac{1}{x} \qquad\text{with $f(0)=100000$}$$
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 $x=0$.
Also $|x^2\sin\frac{1}{x}|\le 1$ for all $x\in (-1,1)$ so that 
$$f(x) =100000+ 100x+ x^2\sin\frac{1}{x} > 100000-100 -1 >0$$ for every $x\in (-1,1).$
A: Define $f(x)=1+3x+x^2\sin\left(\frac1x\right)$ if $x\in(0,1]$ and $f(0)=1$. It satisfies your conditions, but $f'$ is discountinuous at $0$.
A: Take the function $f$ from the question you link. It is already a monotonically increasing continuous function and differentiable on $[a,b]$, yet $f'$ is not continuous. It is only missing positivity.
Make it positive by adding $\lVert f\rVert_\infty+1$ (possible since $f$ is continuous on $[a,b]$, so bounded). This does not change the derivative.
You then have an answer to your new question.
