Differentiability at a point on a compact set implies difference quotients are bounded If $f$ is differentiable at a point $x_0\in [a,b]$ and continuous on $[a,b]$, then $\exists n\in \mathbb{N}$ such that $|f(x)-f(x_0)|\leq n|x-x_0|$ for all $x\in [a,b]$. 
I know that, given that $f$ is differentiable (say $f'(x_0)=c\in \mathbb{R}$), $\displaystyle\left|\lim_{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0}\right|=|c|$. Should I be looking at this limit or at the linear approximation $f(x)-f(x_0)=c(x-x_0)+o(|x-x_0|^2)$? I don't see how either of these imply the claim, even when compactness and continuity are involved. 

You can define the function $g: [a,b]\setminus \{x_0\} \to \Bbb{R}, g(x)=\frac{f(x)-f(x_0)}{x-x_0}$. This function is well defined with one exception: $x_0$. But the existence of the derivative at $x_0$ implies that $g$ has a limit in $x_0$ and can be extended by continuity in $x_0$ by
$$ g_1 : [a,b] \to \Bbb{R},\ g_1(x)=\begin{cases} g(x) & x \neq x_0 \\ f'(x_0) & x=x_0  \end{cases}$$
This function is continuous on a compact set and therefore it is bounded. That boundedness gives the existence of a constant such that $|f(x)-f(x_0)| \leq K|x-x_0|$ (obviously, if $K$ exists, you can choose a bigger positive integer and the inequality holds with an integer constant, if you want it to).
