= Solution
A function is <differentiable> at $a$ when the finite <limit>
$$
f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}h
$$
exists. A derivative need not be continuous. For example,
$$
f(x)=\begin{cases}x^2\sin(1/x),&x\ne0,\\0,&x=0,
\end{cases}
$$
is differentiable everywhere and $f'(0)=0$, but for $x\ne0$,
$$
f'(x)=2x\sin(1/x)-\cos(1/x),
$$
which has no limit at zero.
The <mean value theorem> states that if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then some $c\in(a,b)$ satisfies
$$
f'(c)=\frac{f(b)-f(a)}{b-a}.
$$
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