derivatives - Real Analysis - Uniform continuity

Prove that if $f: (a,b) \rightarrow \Re$ is differentiable on the open interval $(a,b)$, and $f'(x)$ is bounded on the interval $(a,b)$, then $f$ is uniformly continuous on $(a,b)$. Also, prove the converse is false, that is, find a function that is uniformly continuous on $(-1,1)$ whose derivative is unbounded on $(-1,1)$.

So I'm a little confused with how to approach this question but this is what I'm thinking so far.

So for every $\epsilon \gt 0$ there is a $\delta \gt 0$ and there exists $(x,y) \in (a,b)$ such that $|x-y| \lt \delta$... and this is where I get stuck I know it's only the beginning but I don't know how to incorporate the differentiability of $f$ and I have only done uniform continuity on functions such as $1/x$ or others like that can someone guide me in the right direction?

Answer

In general, if $f(x)$ is continuous on [a,b] and $f'(x)$ exists and is bounded by say $M$ on $(a,b)$, then $f(x)$ is Lipschitz continuous(https://en.wikipedia.org/wiki/Lipschitz_continuity), which is a stronger form of uniform continuity.

Indeed, by MVT for any $x,y \in [a,b]$, there exists a $\xi \in (a,b)$ such that (assume $x>y$)

$|f(x)-f(y)| = |(x-y)||f'(\xi)| \leq M|(x-y)|$. Now we let $\epsilon > 0$, and choose $\delta < \frac{\epsilon}{M}$, and we have uniform continuity. In general, MVT is a very powerful tool to bound functions.

For the second part, $f(x) = \frac{1}{x}$ is not uniformly continuous on (0,1) because it can be proven on an open interval (a,b), $f$ is uniformly continuous over (a,b) IF AND ONLY IF it can be extended to a continuous function $g$ such that $g$ is continuous on $[a,b]$ ($g$ will be uniformly continuous over [a,b] as [a,b] is compact).

For a an answer to the second part (A uniformly continuous function such that the derivative is not bounded and is not defined on a compact?).