= Solution
{c}
A <function> $f:(a,b)\to\mathbb R$ is <differentiable> on $(a,b)$ when, at each $c\in(a,b)$, its <difference quotient> has a finite <limit of a function>:
$$
\boxed{f'(c)=\lim_{h\to0}\frac{f(c+h)-f(c)}h\in\mathbb R.}
$$
Here $h\ne0$ and $c+h\in(a,b)$.
For sufficiently small nonzero $h$, the <difference quotient> has <absolute value> at most $|f'(c)|+1$, because it tends to $f'(c)$. Consequently
$$
|f(c+h)-f(c)|\leq|h|\bigl(|f'(c)|+1\bigr)\longrightarrow0.
$$
This is <continuity> at $c$. Since $c$ was arbitrary, \b[<differentiability> implies <continuity> on $(a,b)$.]
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