It's easy to prove that for differentiable functions it is equivalent:
1) has a bounded derivative
2) is Lipschitz
Since Lipschitz unif. cont. we only have to prove has bounded derivative at infinity which follows almost immediately so is unif. cont. in any interval of the form where , but (I believe, I didn't really check it) that it is not unif. cont. in since the sequence seems (!) to map to an unbounded sequence.