Suppose $I$ is an open interval and $f:I\rightarrow\mathbb{R}$ is a differential function. We can say $f$ is uniformly diferentiable if for every $\epsilon> 0$ there exists $\delta> 0$ such that
$x,y\in I$ and $0\lt|x-y|<\delta \Rightarrow \Big| \frac{f(x)-f(y)}{x-y}-f'(x) \Big|\lt\epsilon$
I would like to prove that, if and only if $f'$ is uniformly continuous, then $f$ is uniformly differentiable.
Interchanging $x$ and $y$ we get $\Big| \frac{f(x)-f(y)}{x-y}-f'(y) \Big|\lt\epsilon$ and hence your condition implies $|f'(x)-f'(y)|<2\epsilon$ for all $x,y\in I$ such that $|x-y|<\delta$, or equivalently, $f'$ is uniformly continuous. But there are unifromly continuous functions whose derivative is not uniformly continuous.