= Solution
Assume $f\in C^{n+1}[a,b]$ and reorder the <interpolation nodes> increasingly as $x_0<\cdots<x_n$. The <polynomial interpolation error> $e=f-p_n$ vanishes at all of them. <Rolle's theorem> supplies
$$
\eta_i\in(x_{i-1},x_i),\qquad e'(\eta_i)=0,\qquad i=1,\ldots,n.
$$
These $n$ points are distinct because their containing <intervals> are disjoint. Thus $q=p_n'$ is a <polynomial> of <polynomial degree> at most $n-1$ that interpolates the <function> $f'$ at the <interpolation nodes> $\eta_1,\ldots,\eta_n$.
Apply the <polynomial interpolation error> theorem to $f'$ with these $n$ <interpolation nodes>. Since the $n$th <derivative> of $f'$ is $f^{(n+1)}$, for every $x\in[a,b]$ there is a $\xi_x\in(a,b)$ with
$$
\boxed{f'(x)-q(x)=\frac{f^{(n+1)}(\xi_x)}{n!}\widehat\pi_n(x),\qquad
\widehat\pi_n(x)=\prod_{i=1}^n(x-\eta_i).}
$$
For $x=\eta_i$ the equality is zero on both sides; otherwise the same auxiliary-<function> and repeated-<Rolle's theorem> proof applies. This also covers $n=1$, where $q$ is a <constant polynomial> and the proof is the ordinary <mean value theorem> for $f'$.
Writing $M=\max_{[a,b]}|f^{(n+1)}|$ gives the genuine error bound
$$
\boxed{|f'(x)-q(x)|\leq\frac M{n!}\prod_{i=1}^n|x-\eta_i|
\leq\frac M{n!}(b-a)^n.}
$$
The first formula is sharper, but its intermediate <interpolation nodes> depend on $f$; the last bound eliminates those unknown locations. This is <derivative interpolation error from Rolle points>.
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