Taylor series of $f(x)=\frac{1}{4+x^2}$, with $f:D\to\mathbb{R}, D\subseteq \mathbb{R}$ at $x^*=0$ and determining the radius of convergence of the series.
I've calculated the first 3 derivatives in order to figure out the $n$-th derivative. Unfortunately I couldn't really find a pattern. In order to create the series I will need the $n$-th derivative, that's why I'm stuck here. Can you help me with finding the $n$-th derivative?
$\begin{align} f(x)'=&-\frac{2x}{(4+x^2)^2}\\ f(x)''=&\frac{6x^2-8}{(x^2+4)^3}=\dfrac{2\left(\color{limegreen}{3x^2-4}\right)}{\left(x^2+4\right)^3}\\ f(x)'''=&-\frac{24x(\color{blue}{x^2-4})}{(x^2+4)^4}=\dfrac{12x}{\left(x^2+4\right)^3}-\dfrac{12x\left(\color{limegreen}{3x^2-4}\right)}{\left(x^2+4\right)^4}\\ f(x)^{(4)}=&\frac{24\left(\color{red}{5x^4-40x^2+16}\right)}{\left(x^2+4\right)^5}=-\frac{48x^2}{\left(x^2+4\right)^4}-\frac{24\left(\color{blue}{x^2-4}\right)}{\left(x^2+4\right)^4}+\frac{192x^2\left(x^2-4\right)}{\left(x^2+4\right)^5}\\ f(x)^{(5)}=&-\frac{240x\left(3x^4-40x^2+48\right)}{\left(x^2+4\right)^6}=\frac{24\left(20x^3-80x\right)}{\left(x^2+4\right)^5}-\frac{240x\left(\color{red}{5x^4-40x^2+16}\right)}{\left(x^2+4\right)^6} \end{align}$
I can see some sort of pattern highlighted in different colors but I'm not sure how I could express it exactly.
It should look similar to this:
$$f^{(n)}(x)=(-1)^n\cdot \frac{?}{(4+x^2)^n}$$