TITLE: Lipschitz implies bounded derivative?
QUESTION [6 upvotes]: Is there any function which is Lipschitz but has an unbounded derivative ? I guess there isn't but I have not found one.

REPLY [10 votes]: By definition of Lipschitz, $f'$ is necessarily bounded since $|f(x) -f(y)|\le L |x - y| $ which implies that 
$$|g'(x)| = \left|\lim_{y\to x} \frac{f(y) - f(x)}{y-x} \right| \le L$$
for all  $x$ (assuming $g'(x)$ exists).