real analysis - If $f$ is uniformly differentiable $Longrightarrow $ $f'$ is continuous

I'm not sure how to go about proving this theorem:

Let $U\subset \mathbb{R}^m$ (open set) and $f:U\longrightarrow \mathbb{R}^n$ a differentiable function such that:

$\forall \epsilon>0\,,\exists \delta>0:|\!|h|\!|<\delta,[x,x+h]\subset U \Longrightarrow |\!|f(x+h)-f(x)-f'(x)(h)|\!|<\epsilon |\!|h|\!|$

then it holds that $f':U\longrightarrow \mathcal{L}(\mathbb{R}^m,\mathbb{R}^n)$ is continuous.We can also say that $f'$ is uniformly continuous?

Any hints would be appreciated.

Answer

If you interchange the roles of $x$ and $x+h$ (and replace $h$ by $-h$), then you see that
$$
|f(x)-f(x+h)+f'(x+h)(h)|<\epsilon|h|,
$$
which, upon combining with what you have, and using the triangle inequality, shows that $f'(x)$ and $f'(x+h)$ are $\epsilon$-close.