Consider $f: U \to \mathbb{R}^m$, where $U$ is an open ball in $\mathbb{R}^n$.
Suppose $f$ is uniformly differentiable on $U$:
$$\forall \epsilon>0\,,\exists \delta>0:|\!|{\bf h}|\!|<\delta,{\bf x},{\bf x}+{\bf h}\in U \Longrightarrow |\!|f({\bf x}+{\bf h})-f({\bf x})-f'({\bf x})({\bf h})|\!|<\epsilon \|{\bf h}\|$$
That is, the $\delta$ is valid for all ${\bf x}$.
I wonder if the following is true:
Question: Can we always construct an extension of $\bar{f}$ of $f$, such that $\bar{f}({\bf x}) = f({\bf x})$, for ${\bf x} \in U$, and $\bar{f}$ uniformly differentiable on $\mathbb{R}^n$?
Consider $n=m=1$, $U=(-1,1)$, $f(x)=\sqrt{x+1}$.
Obviously $f$ is differentiable on $U$, but not at $x=-1$.
In fact, the tangent there would become vertical, or equivalently:
$f'(x)$ would increase beyond any bound.
--- rk