Finding the n-th derivative of a rational function

Question

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.

Answer 1

Using partial fractions you should get $$f(x)=A(x-a)^{-1}+B(x-b)^{-1}$$

Then $n$th derivative of $(x-c)^{-1}$ is easy to calculate.

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Note: $$\frac{1}{1-x-x^2}=\left(\frac{1}{\sqrt{5}}\right)\left(\frac{1}{x+\frac{\sqrt{5}+1}{2}}\right)-\left(\frac{1}{\sqrt{5}}\right)\left(\frac{1}{x-\frac{\sqrt{5}-1}{2}}\right)$$

Answer 2

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