Finding the n-th derivative of a rational function Given $$f : \left(\frac{-1}{2}, \frac{1}{2}\right) \to \mathbb{R} \ \ \ \ \ \ 
 \ \ \ \ f(x) = \frac{1}{1-x-x^2}$$
I am trying to find an expression for $f^{n}(x)$ and prove it by induction afterwards. 
I tried first splitting into partial fractions but it got messy, not sure if that's the right approach. 

$$
1-x-x^2 = -\left(x-\frac{-1+\sqrt 5} 2\right)\left(x-\frac{-1-\sqrt5} 2\right) = \left( \frac{-1+\sqrt 5} 2 - x \right) \left( \frac{1+\sqrt 5} 2 + x \right)
$$
$$
\frac A {\frac{-1+\sqrt 5} 2 - x} + \frac B {\frac{1+\sqrt5} 2 + x} = \frac 1{\left( \frac{-1+\sqrt 5} 2 - x \right) \left( \frac{1+\sqrt 5} 2 + x \right)}
$$
$$
A\left(\frac{1+\sqrt 5} 2 + x \right) + B\left( \frac{-1+\sqrt 5} 2 - x \right) = 0x + 1
$$
So $A-B=0$ and $(A+B) \left( \dfrac{\sqrt 5} 2 \right) = 1.$ Hence $A=B=\dfrac 1 {\sqrt 5}.$
