Find $f^{(100)}(x)$ where $f(x)=\frac{1}{4x^2-1}$ Find $f^{(100)}(x)$ where $f(x)=\frac{1}{4x^2-1}$. I found first,second and third derivative: 
$$f'(x)=\frac{-8x}{(4x^2-1)^2}$$
$$f''(x)=\frac{96x^2+8}{(4x^2-1)^3} $$
$$f'''(x)=\frac{-384x(4x^2+1)}{(4x^2-1)^4}$$
I can't seem to find any rule between them. Anyone has any ideas or hints?

$$
f(x) =  \frac{1}{2}\left(\frac{1}{2x-1} - \frac{1}{2x+1}\right)
$$
Now if $g(x) = \frac{1}{2x-1} = (2x-1)^{-1}$, then check (by induction) that
$$
g^{(n)}(x) = (-2)^nn!(2x-1)^{-(n+1)}
$$
and if $h(x)=(2x+1)^{-1}$, then
$$
h^{(n)}(x) = (-2)^nn!(2x+1)^{-(n+1)}
$$
Hence,
$$
f^{(n)}(x) = \frac{(-2)^nn!}{2} \left(\frac{1}{(2x-1)^{(n+1)}} - \frac{1}{(2x+1)^{(n+1)}}\right)
$$
Now take $n=100$.
