I am trying to derive the error term for the backward differentiation formula $f'(x) \approx \frac{3f(x)-4f(x-h)+f(x-2h)}{2h}$.
By Taylor expansion
$f(x-h)=f(x)-hf'(x)+\frac{h^2}{2}f''(x)-\frac{h^3}{6}f^3(\xi_a)$,
$f(x-h)=f(x)-2hf'(x)+2h^2f''(x)-\frac{4h^3}{3}f^3(\xi_b)$.
So in forming the linear combination I get $\frac{3f(x)-4f(x-h)+f(x-2h)}{2h}=f'(x)+\frac{h^2}{3}f^3(\xi_a)-\frac{2h^2}{3}f^3(\xi_b)\leq f'(x)+\frac{2h^2}{3}|f^3(\xi)|$, where $|f^3(x_0)|\leq|f^3(\xi)|$ for all $x_0\in(x,x-2h)$.
So $\frac{2h^2}{3}|f^3(\xi)|$ is the error bound. But the notes I am following, which uses an identical method, ends up with $\frac{16}{3}$ as the coefficient for $h^2$ in the error formula (i.e. $\frac{16h^2}{3}|f^3(\xi)|$ is the error bound). Where am I going wrong?
The expected error is second order, so consider and treat with the extended mean value theorem \begin{align} &\frac{3f(x)-4f(x-h)+f(x-2h)-2hf'(x)}{2h^3} \\[.5em] &=\frac{4f'(x-h_1)-2f'(x-2h_1)-2f'(x)}{6h_1^2} \\[.5em] &=\frac{-4f''(x-h_2)+4f''(x-2h_2)}{12h_2} \\[.5em] &=-\frac{f'''(x-h_3)}3 \end{align} where $0<h_2<h_1<h$ and $h_3\in[h_2,2h_2]$, so that finally $$ \frac{3f(x)-4f(x-h)+f(x-2h)}{2h}=f'(x)-\frac{h^2}3f'''(\xi), ~~ \xi\in[x-2h,x] $$ This means that the error term is even smaller than the one that you derived.