real analysis - Is there a function on a compact interval that is differentiable but not Lipschitz continuous?

Consider a function $f:[a,b]\rightarrow \mathbb{R}$, does there exist a differentiable function that is not Lipschitz continuous?

After discussing this with friends we have come to the conclusion that none exist. However there is every chance we are wrong. If it is true that none exist how could we go about proving that? It is true that if $f$ is continuously differentiable then $f$ is Lipschitz, but what if we don't assume the derivative is continuous?

Answer

The map $f : [0,1] \to \mathbb{R}$, $f(0) = 0$ and $f(x) = x^{3/2} \sin(1/x)$ is differentiable on $[0,1]$ (in particular $f'(0) = \lim_{x \to 0^+} f(x)/x = 0$), but it is not Lipschitz (the derivative $f'(x)$ is unbounded).