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The error of interpolating polynomial is $$ E_n(x)=\frac{(x-x_0)(x-x_1)\cdots(x-x_n)}{(n+1)!}f^{(n+1)}(\xi(x)) $$ The derivative of $E_n(x)$ is $$ E^{\prime}_n(x)=\frac{d}{dx}\left((x-x_0)(x-x_1)\cdots(x-x_n)\right)\frac{f^{(n+1)}(\xi(x))}{(n+1)!}\\ +(x-x_0)(x-x_1)\cdots(x-x_n)\frac{d}{dx}\left(\frac{f^{(n+1)}(\xi(x))}{(n+1)!}\right) $$ In A Friendly Introduction to Numerical Analysis by Brian Bradie, the author gives the derivative of $f^{(3)}(\xi(x))/6$ at $x=x_0$ in the terms of divided differences $$ \frac{d}{dx}\left(\frac{f^{(3)}(\xi(x))}{6}\right)_{\text{at }x=x_0}=\frac{d}{dx}f[x_0-h,x_0,x_0+h,x]_{\text{at }x=x_0}\\ $$ $$ \begin{align} & =\lim_{\Delta\to0}\frac{f[x_0-h,x_0,x_0+h,x_0+\Delta]-f[x_0-h,x_0,x_0+h,x_0]}{\Delta}\\ & =\lim_{\Delta\to0}f[x_0-h,x_0,x_0+h,x_0+\Delta,x_0]\\ & =\lim_{\Delta\to0}f[x_0-h,x_0,x_0+h,x_0,x_0]\\ & =\frac{1}{24}f^{(4)}(\xi(x))\\ \end{align} $$

But I don't understand this manner. How can I obtain $\frac{1}{24}f^{(4)}(\xi(x))$ without using divided differences?

Hint: $$ \frac{1}{(n+1)!}\frac{d^i}{dx^i}f^{(n+1)}(\xi)=\frac{i!}{(n+i+1)!}f^{(n+i+1)}(\rho_i) $$ where $i=1,2,\cdots,n$ and $\min(x_0,x_1,\cdots,x_n,x)<\rho_i<\max(x_0,x_1,\cdots,x_n,x)$.

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