$f(x) = \frac{4 + x}{2 + x - x^2}$, calculate $f^{(9)}(1)$ $f(x) = \frac{4 + x}{2 + x - x^2}$, calculate $f^{(9)}(1)$, where $f^{(9)}$ is the $9$-th derivative of $f$.
Domain of $f$ is $\mathbb{R} - \{-1, 2\}$. I've got that $f(x) = \frac{1}{1 - (-x)} + \frac{1}{1 - \frac{1}{2}x} = \sum_{n=0}^\infty ((-1)^n + 2^{-n})x^n$, but there is a problem that $\frac{1}{1 - (-x)} = \sum_{n=0}^\infty (-1)^nx^n$ is convergent only for $|x| < 1$, so not for $1$. How can I go about this?

I've solved this using Julian Mejia's suggestion by making substitution $t = x -1$. We have $g(t) = \frac{1}{2} \cdot \left(\frac{4}{1 - t} + \frac{1}{1 + \frac{1}{2}t}\right) = \sum_{n=0}^\infty (2-(-\frac{1}{2})^{n+1})t^n$ which converges for $t \in (-1, 1)$. Now $9$-th derivative is just $(2-(-\frac{1}{2})^{10}) \cdot 9!$, which is exactly what InterstellarProbe got.
