If $f$ is differentiable and its derivative is uniformly continuous, then $n \cdot (f(x+\frac{1}{n})-f(x)) \to^{u} f'(x)$ Let $f$ be a differentiable function in $\mathbb{R}$ such that its derivative $f'$ is uniformly continuous in $\mathbb{R}$. 
Prove: $$n \cdot (f(x+\frac{1}{n})-f(x)) \to^{u} f'(x)$$ (uniform convergence).
If $f'$ is only continuous- Does the claim still hold?
Please help me approach this one

Instead of using integration, you can use the mean value theorem as suggested in the comments. 
Let $\epsilon > 0$, so we have $\delta > 0$ such that $|x-y|< \delta$, $|f'(x) - f'(y)| < \epsilon$. Now let $n > \frac{1}\delta$ and $x \in \mathbb R$. By the mean value theorem, $\exists \xi \in (x, x+\frac{1}{n})$ such that 
$$\frac{f\left(x+\frac{1}{n}\right) - f(x)}{\frac{1}{n}}=f'(\xi).$$
Since $n > \frac{1}{\delta}$, $|\xi - x| < \delta$, so $|f'(\xi) - f'(x)|<\epsilon$. Combining these two results,
$$\left|\frac{f\left(x+\frac{1}{n}\right) - f(x)}{\frac{1}{n}}-f'(x)\right|=\left|f'(\xi)-f'(x)\right|<\epsilon,$$
as required.
