A differentiable function on an interval is Lipschitz exactly when its derivative is bounded, and the best constant is $\sup\abs{f^\prime}$.Suggest a slogan
Let \(I\subseteq\RR\) be an interval with nonempty interior and let \(f\colon I\to\RR\) be differentiable. Then \(f\) is Lipschitz, that is, there is \(C\geq0\) with \(\abs{f(x)-f(y)}\leq C\abs{x-y}\) for all \(x,y\in I\), if and only if \(f'\) is bounded on \(I\). In that case the smallest such constant is \(C=\sup_{x\in I}\abs{f'(x)}\).
A Lipschitz function need not be differentiable: \(x\mapsto\abs{x}\) on \((-1,1)\) satisfies \(\abs{\abs x-\abs y}\leq\abs{x-y}\) and is not differentiable at \(0\).
The interval hypothesis is needed: on \(U\coloneqq(0,1)\cup(1,2)\), the function \(f\coloneqq\chi_{(1,2)}\) is differentiable with \(f'=0\), but \(\abs{f(1+\delta)-f(1-\delta)}=1\) for \(0<\delta<1\), so \(f\) is not Lipschitz on \(U\).