Finite difference formulas for f′

1. $f(x\pm h)=f\pm hf'+\frac{h^2}2f''\pm\frac{h^3}6f'''+\frac{h^4}{24}f^{(4)}\pm\frac{h^5}{120}f^{(5)}+\cdots$, so
$$f(x+h)-f(x-h)=2hf'+\frac{h^3}3f'''+\frac{h^5}{60}f^{(5)}+\cdots$$
Dividing by $2h$ gives the first formula.

2. Let $D(h)=\dfrac{f(x+h)-f(x-h)}{2h}=f'+\dfrac{h^2}6f'''+\dfrac{h^4}{120}f^{(5)}+\cdots$ (from part 1, pushed one order further). Consider $\dfrac{4D(h)-D(2h)}3$:
- the coefficient of $f'$ is $\frac{4-1}3=1$,
- the coefficient of $h^2f'''$ is $\frac{4/6-4/6}3=0$,
- the coefficient of $h^4f^{(5)}$ is $\frac{4/120-16/120}3=-\frac1{30}$.

Explicitly,
$$\frac{4D(h)-D(2h)}3=\frac{8\left(f(x+h)-f(x-h)\right)-\left(f(x+2h)-f(x-2h)\right)}{12h},$$
which is the stated formula, with error $-\frac{h^4}{30}f^{(5)}(x)$.

Numerics. For $f=\exp$, $x=0$, $h=0.1$ (exact value $1$):
- formula 1 gives $1.0016675$ (error $1.7\cdot10^{-3}\approx\frac{h^2}6$),
- formula 2 gives $0.9999967$ (error $-3.3\cdot10^{-6}\approx-\frac{h^4}{30}$).

Remark. Cancelling terms through a clever combination gains two orders of accuracy for the price of two more function evaluations.