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 you don't know the theory of generating functions, the trick is to get the Taylor expansion with partial fractions. Write
$$
\frac{1}{1-x-x^2}=\frac{a}{1-\alpha x}+\frac{b}{1-\beta x}
$$
that gives the relations
$$
\begin{cases}
a+b=1\\[4px]
a\beta+b\alpha=0\\[4px]
\alpha+\beta=1\\[4px]
\alpha\beta=-1
\end{cases}
$$
Now you can use
$$
\frac{a}{1-\alpha x}=a\sum_{n\ge0}\alpha^nx^n
$$
so
$$
\frac{f^{(10)}(0)}{10!}=a\alpha^{10}+b\beta^{10}
$$

Once you compute $a$, $b$, $\alpha$ and $\beta$, you'll notice that
$$
a\alpha^n+b\beta^n=\frac{\varphi^n-\bar{\varphi}^n}{\sqrt{5}}
$$
where
$$
\varphi=\frac{1+\sqrt{5}}{2},\qquad
\bar{\varphi}=\frac{1-\sqrt{5}}{2}
$$
which is the Bézout formula for the Fibonacci numbers.
