Quadratic lagrange interpolation is given by $P(x)=\sum _{k=0}^2 l_k(x) f_k$ with error term $E(x)=\frac{(x-x_0)(x-x_1)(x-x_2) f^{'''}(\xi )}{3!}$. To calculate, error for second derivative at point any point, first I calculated $$E'(x)=f^{'''}(\xi)\frac{(x-x_0)(x-x_1)+(x-x_0)(x-x_2)+(x-x_2)(x-x_1)}{3!}+\frac{(x-x_0)(x-x_1)(x-x_2)}{3!}\frac{d}{dx}f^{'''}(\xi),$$ where $x_0<\xi<x_2$ then $$E''(x)=f^{'''}(\xi)\frac{(2x-x_0-x_1)+(2x-x_0-x_2)+(2x-x_2-x_1)}{3!}+ \frac{(x-x_0)(x-x_1)+(x-x_0)(x-x_2)+(x-x_2)(x-x_1)}{3!}\frac{d}{dx}f^{'''}(\xi)+\frac{(x-x_0)(x-x_1)+(x-x_0)(x-x_2)+(x-x_2)(x-x_1)}{3!}\frac{d}{dx}f^{'''}(\xi)+ \frac{(x-x_0)(x-x_1)(x-x_2)}{3!}\frac{d^2}{dx^2}f^{'''}(\xi)$$
Therefore at $x=x_0$, $E^{''}(x_0)= f^{'''}(\xi)\frac{(2x_0-x_1-x_2)}{3}+\frac{(x_0-x_1)(x_0-x_2)}{6}\frac{d}{dx}f^{'''}(\xi)+\frac{(x_0-x_1)(x_0-x_2)}{6}\frac{d}{dx}f^{'''}(\xi)$.
I need to show that this value is equal to $E^{''}(x_0)= f^{'''}(\xi)\frac{(2x_0-x_1-x_2)}{3}+\frac{(x_0-x_1)(x_0-x_2)}{24}[f^{iv}(\eta_1)+f^{iv}(\eta_2)]$.
Any help is appreciated.