Prove: bounded derivative if and only if uniform continuity

As Jose27 noted, uniformly continuous functions need not be differentiable even at a single point.

It is true that if $f$ is defined on an interval in $\mathbb R$ and is everywhere differentiable with bounded derivative, then $f$ is uniformly continuous. In fact, it follows from the Mean Value Theorem that such an $f$ is Lipschitz, which is much stronger.

However, if $f$ is uniformly continuous and everywhere differentiable, then $f$ need not have bounded derivative. Jose27 mentions $\sqrt x$, which would work as an example on the interval $(0,\infty)$. The function $$f\left(x\right)=\begin{cases}x^2\sin\left(\frac{1}{x^2}\right) & :x\neq 0\\ 0 &:x=0\end{cases}$$ is uniformly continuous on any bounded interval such as $(-1,1)$, but has unbounded derivative near $0$. There are also examples where $f'$ is bounded on bounded intervals, but unbounded on $\mathbb R$, while $f$ is uniformly continuous. You can show that any continuous function $f$ on $\mathbb R$ such that $\lim\limits_{|x|\to \infty}f(x)=0$ is uniformly continuous, and using this fact you can see that Nate Eldredge's example here of $\sin(x^4)/(1+x^2)$ provides such an example. Another source of examples is the question

Why if $f'$ is unbounded, then $f$ isn't uniformly continuous?