Derivative interpolation error from Rolle points ($f^{\prime }(x)-p_n^{\prime }(x)=\frac{f^{(n+1)}(\xi )}{n!}{\prod }_{i=1}^n(x-{\eta }_i)$)

Let $f\in C^{n+1}[a,b]$, $n\ge 1$, 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 $f^{\prime }({\eta }_i)=p_n^{\prime }({\eta }_i)$. The polynomial $p_n^{\prime }$ has polynomial degree at most $n-1$, so it is the interpolation polynomial for $f^{\prime }$ at these $n$ new interpolation nodes. Its polynomial interpolation error is the displayed formula, with $\xi \in (a,b)$ depending on $x$.

Consequently $|f^{\prime }(x)-p_n^{\prime }(x)|\le M{\prod }_i|x-{\eta }_i|/n!\le M(b-a{)}^n/n!$, 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.