If f is uniformly differentialable on $(a,b)$ then $f'$ is continuous on $(a,b)$?

Question

If f is uniformly differentialable on $(a,b)$ then $f'$ is continuous on $(a,b)$?

I just knew the definition

A function $f:(a,b)\to R$ is said to be uniformly differentialable iff f is differntiable on $(a,b)$, and for each $\epsilon>0$ there is $\delta>0$ such that $0<|x-y|<\delta$ and for any $x,y \in (a,b)$ imply that $\left|\frac{f(x)-f(y)}{x-y}-f'(x)\right|<\epsilon$

Is that this statement is true :

f is uniformly differentialable on $(a,b)$ then $f'$ is continuous on $(a,b)$?

Answer 1

Given $x \in (a,b)$ \begin{align*} |f'(y)-f'(x)| & \leq |f'(x) - \frac{f(x)-f(y)}{x-y} | + |\frac{f(x)-f(y)}{x-y}-f'(y)| \end{align*} Given that $f$ is differentiable at $x$, the first member of the sum is small for $y$ sufficiently close to $x$. The second member is small for $y$ sufficiently close to $x$, for $f$ is uniformly differentiable on $(a,b)$.

Answer 2

Hint:

For $x < z < y$, $|x-y| < \delta \implies |x-z|, |y-z| < \delta$ and

$$f'(x) - f'(y) \\ = f'(x) - \frac{f(x) - f(z)}{x-z} + \frac{f(x) - f(z)}{x-z} - f'(z) + f'(z) - \frac{f(z) - f(y)}{z-y} + \frac{f(z) - f(y)}{z-y} - f'(y) $$