If $f(x)=\frac{1}{1-x-x^2}$, find $\frac{f^{(10)}(0)}{10!}$ My math teacher posed this question to my calculus class. If $f(x)=\frac{1}{1-x-x^2}$, find $\frac{f^{(10)}(0)}{10!}$.
At first, I began by taking the first few derivatives, but it soon go out of hand with the repeated quotient rules. I'm sure that I could continue for $10$ derivatives, but I believe that there must be an easier solution.

If we consider 
$$ g(z) = \frac{z}{1-z-z^2} = \sum_{n\geq 0} a_n z^n $$
we have an analytic function in a neighbourhood of zero whose coefficients fulfill $a_0=0, a_1=1$ and $a_{n+2}-a_{n+1}-a_{n}=0$ for any $n\geq 0$ (it is enough to compute the coefficient of $z^{n+2}$ in $z=(1-z-z^2)\sum_{n\geq 0}a_n z^n$). It follows that $a_n = F_n$, the $n$-th Fibonacci number, so:
$$ f(z) = \frac{g(z)}{z} = \frac{1}{1-z-z^2} = \sum_{n\geq 0}\frac{f^{(n)}(0)}{n!}\cdot z^n = \sum_{n\geq 0} F_{n+1}\cdot z^n $$
and the answer is given by $\color{red}{F_{11}=89}$.
