Originally Posted by

**Laurent** The question asks to show that "differentiable" implies "Lipschitz", not the converse (which is false) (however, a Lipschitz function is differentiable almost-everywhere, but this is a hard theorem).

To solve the question, you just have to write the $\displaystyle \varepsilon-\delta$ definition of the differentiability at $\displaystyle x$ for a fixed $\displaystyle \delta$, for instance $\displaystyle \delta=1$, and to use a triangle inequality: there is $\displaystyle \ell(=f'(x))$ and $\displaystyle \varepsilon>0$ such that for y such that $\displaystyle |x-y|<\varepsilon$, $\displaystyle |\frac{f(y)-f(x)}{y-x}-\ell|\leq 1$, hence $\displaystyle |\frac{f(y)-f(x)}{y-x}|\leq \ell + 1$.