TITLE: Uniform Continuity and Differentiation
QUESTION [5 upvotes]: Is the following true or false?: 

Let $f\colon  [0,1) \to \mathbb{R}$ be a function differentiable in $[0,1)$ (where the derivative at zero means "right derivative") such that both $f$ and $f'$ are uniformly continuous in $(0,1)$. Then $f'$ is continuous.

Note that the mistery lies at $x=0$. So the question is: can we say with these hypotheses that $f'(0)=\lim_{x\to 0^{+}}f'(x)$ (which exists thanks to the uniform continuity of $f'_{\mid (0,1)}$). Note also that the uniform continuity of $f'_{\mid (0,1)}$  makes redundant the analogous requirement for $f$ (which will even more become a Lipschitz function).

REPLY [2 votes]: As you stated, thanks to the uniform continuity of $f'$, $L=\lim\limits_{x\rightarrow0^+} f'(x)$ exists (uniformly continuous functions map Cauchy sequences to Cauchy sequences).
If $L\ne f'(0)$, then a contradiction to Darboux's Theorem can be obtained.
Darboux's Theorem: If $f$ is differentiable on $I=[a,b]$ and if $k$ is a number between $f'(a)$ and $f'(b)$, then there is at least one point $c\in(a,b)$ such that $f'(c)=k$.