= Derivative interpolation error from Rolle points
{title2=$f'(x)-p_n'(x)=\dfrac{f^{(n+1)}(\xi)}{n!}\prod_{i=1}^n(x-\eta_i)$}
Let $f\in C^{n+1}[a,b]$, $n\geq1$, and let $p_n$ be its <Lagrange interpolation polynomial> at ordered distinct <interpolation nodes> $x_0<\cdots<x_n$. Since $f-p_n$ vanishes at every <interpolation node>, <Rolle's theorem> supplies distinct $\eta_i\in(x_{i-1},x_i)$ where $f'(\eta_i)=p_n'(\eta_i)$. The <polynomial> $p_n'$ has <polynomial degree> at most $n-1$, so it is the <interpolation polynomial> for $f'$ at these $n$ new <interpolation nodes>. Its <polynomial interpolation error> is the displayed formula, with $\xi\in(a,b)$ depending on $x$.
Consequently $|f'(x)-p_n'(x)|\leq M\prod_i|x-\eta_i|/n!\leq M(b-a)^n/n!$, where $M=\max|f^{(n+1)}|$. The existence of the <interpolation nodes> is enough for this bound; their explicit locations generally depend on $f$. This proof avoids differentiating an unspecified intermediate point in the original <polynomial interpolation error> formula.
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