<s>[INST] find $f^{(15)}(0)$ if $f = \frac{10x^2+12x+4}{(x+2) (x^2 +1)}$ $$f = \frac{10x^2+12x+4}{(x+2)(x^2 +1)}=\frac{4}{x+2}+\frac{6x}{x^2+1}=\frac{2}{1-\left(-\frac x 2 \right)}+\frac{6x}{1-(-x^2)}$$
So we end up with $$2 \sum_{n=0}^{15} \left(-\frac x 2 \right)^n + 6x \sum_{n=0}^{15} (-x^2)^n$$ but because we are looking for $f^{(15)}(0)$ we get the result of $2$?
 [/INST] We have
\begin{align*}
f(x) = \frac{4}{x+2} + \frac{3}{x+i} + \frac{3}{x-i}
\end{align*}
and hence
\begin{align*}
f^{15}(x) = -(15!)(4(x+2)^{-16} + 3(x+i)^{-16} + 3(x-i)^{-16}
\end{align*}
and 
\begin{align*}
f^{15}(0) = -(2^{-14}+6)\cdot 15!
\end{align*}
</s>