泰勒展开证明导数公式 f'(x) = [f(x-2h) - 8f(x-h) + 8f(x+h) - f(x+2h)] / 12h + O(h^4)

根据泰勒展开公式：

f(x ± h) = f(x) ± hf'(x) + h^2/2!f''(x) ± h^3/3!f'''(x) + h^4/4!f^{(4)}(x) + O(h^5)

将其代入式子：

f'(x) = (f(x+h)-f(x))/h - h/2f''(x) + h^2/6f'''(x) - h^3/24f^{(4)}(x) + O(h^4)

f'(x) = (f(x+2h)-f(x))/2h - h/2f''(x) + 2h^2/3!f'''(x) - 2h^3/4!f^{(4)}(x) + O(h^4)

f'(x) = (8f(x+h)-8f(x-h))/4h + O(h^3)

将第二个式子乘以3和第三个式子乘以-1，然后相加可得：

3f'(x) = (3f(x+2h)-3f(x))/2h - 3hf''(x) + 3 * 2h^2/3!f'''(x) - 3 * 2h^3/4!f^{(4)}(x) + O(h^4)

-f'(x) = (-f(x-2h)+f(x))/2h + hf''(x) + h^2/3!f'''(x) + h^3/4!f^{(4)}(x) + O(h^4)

将两式相加可得：

2f'(x) = (f(x+2h)-f(x-2h))/2h - 8f'(x) + 8 * 2h^2/3!f'''(x) - 2 * 2h^3/4!f^{(4)}(x) + O(h^4)

整理得：

f'(x) = (f(x+2h)-8f(x+h)+8f(x-h)-f(x-2h))/12h + O(h^4)

即：

f'(x) = (f(x-2h)-8f(x-h)+8f(x+h)-f(x+2h))/12h + O(h^4)

证毕。