TITLE: What is the $n^\text{th}$ derivative of $f(x)=\frac{1}{1+x^2}$
QUESTION [5 upvotes]: I want the taylor series expansion around some value $a$ of the function $f(x)=\frac{1}{1+x^2}$. I used the general formula
\begin{eqnarray}
f(x) = f(a) + \sum_{n=1}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n
\end{eqnarray}
But unfortunately, I cannot compute any general formula for $f^{(n)}(a)$. The first derivative is 
$$ f^{(1)}(x)= -\frac{2x}{(1+x^2)^2}.$$The second derivative is 
$$ f^{(2)}(x)= \frac{6x^2-2}{(1+x^2)^3}.$$The third derivative is 
$$ f^{(3)}(x)= \frac{24x(x^2-1)}{(1+x^2)^4}.$$The fourth derivative is 
$$ f^{(4)}(x)= -\frac{24(5x^4-10x^2+12)}{(1+x^2)^5}$$. The fifth derivative is 
$$ f^{(5)}(x)= \frac{240x(3x^4-10x^2+3)}{(1+x^2)^5}$$.
What is the $n$-th derivative of the function for working with the above taylor series which I want to use to prove something?

REPLY [5 votes]: Another form of the $n$-th derivative:
$$f^{(n)}(x)=(-1)^n n!\frac{\sin((n+1)\cot^{-1} x)}{(1+x^2)^{(n+1)/2}}.$$
It's easy to prove by induction.