show that if $f$ is uniformly differentiable then prove that $f'$ is continuous How would I go about showing that if $f$ is uniformly differentiable then $f'$ is continuous.
my attempt:
A differentiable function $f:[a,b]\to \Bbb R$ is said to be uniformly differentiable on $[a,b]$ if $\forall \epsilon>0 \exists \delta>0:\forall x,y\in[a,b]$ we have $$0<|x-y|<\delta\implies |\frac{f(x)-f(y)}{x-y}-f'(y)|<\epsilon$$
I don't know where to go from here, any help would be highly appreciated.
 A: Let $\varepsilon > 0$. Since $f$ is uniformly differentiable there exists a $\delta > 0$ such that $\forall x,y\in[a,b]$ if $|x-y| < \delta$ then:
$$
\bigg{|}\frac{f(x)-f(y)}{x-y} - f'(x)\bigg{|} < \frac{\varepsilon}{2}
$$
interchanging $x$ and $y$ (why can you do this?) you get:
$$
\bigg{|}\frac{f(x)-f(y)}{x-y} - f'(y)\bigg{|} < \frac{\varepsilon}{2}
$$
Now let $|x-y|< \delta$ then (adding and substracting the same term):
$$
|f'(x)-f'(y)| = \bigg{|} f'(x) + \frac{f(x)-f(y)}{x-y} - \frac{f(x)-f(y)}{x-y} - f'(y) \bigg{|} \\
\leq\bigg{|} \frac{f(x)-f(y)}{x-y} - f'(x)\bigg{|} + \bigg{|} \frac{f(x)-f(y)}{x-y} - f'(y) \bigg{|} \\
< \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon
$$
A: The answers from Ethan and Juan work, but here is a more conceptual approach. 
Write $F_n(y)=\frac{f(y+\frac{1}{n})-f(y)}{\frac{1}{n}}$, then the assumption of uniform differentiability implies that 
$$F_n(y) \text{  converges to  }f'(y)\text{ uniformly on }[a,b].$$
Note that the the $F_n$'s are all continuous functions. Then by the fact that uniform limit of continuous functions is also continuous, $f'(y)$ is continuous. 
A: HINT:
$$\vert f'(x) - f'(y)\vert \leq \left\vert f'(x) - \frac{f(x) - f(y)}{x - y} \right\vert + \left\vert \frac{f(x) - f(y)}{x - y} - f'(y)\right\vert$$
