Good day everyone.

I'm so stuck with this homework in numerical methods. I'm supposed to derive an equation using Taylor expansion. The details are below:

Use the Taylor expansions for f(x+h), f(x-h), f(x+2h) and f(x-2h) and derive the central-difference formula:

f^{(4)}(x)=\frac{f(x+2h)-4f(x+h)+6f(x)-4f(x-h)+f(x-2h)}{h^4}


I know I'm supposed to start with these four equations,

f(x+h)=f(x)+f'(x)h+\frac{f^{(2)}(x)h^2}{2!} + \frac{f^{(3)}(x)h^3}{3!} + \frac{f^{(4)}(x)h^4}{4!} + \frac{f^{(5)}(x)h^5}{5!}
f(x+h)=f(x)-f'(x)h+\frac{f^{(2)}(x)h^2}{2!} -\frac{f^{(3)}(x)h^3}{3!} + \frac{f^{(4)}(x)h^4}{4!} - \frac{f^{(5)}(x)h^5}{5!}

f(x+2h)=f(x)+f'(x)2h+\frac{f^{(2)}(x)4h^2}{2!} + \frac{f^{(3)}(x)8h^3}{3!} + \frac{f^{(4)}(x)16h^4}{4!} + \frac{f^{(5)}(x)32h^5}{5!}
f(x-2h)=f(x)-f'(x)2h+\frac{f^{(2)}(x)4h^2}{2!} - \frac{f^{(3)}(x)8h^3}{3!} + \frac{f^{(4)}(x)16h^4}{4!} - \frac{f^{(5)}(x)32h^5}{5!}

After these, I can't move on. HELP!