# Nth derivative function

Is there any technique to find the $n$th derivative of $1/(1+x^2)$? I have been trying to find the $n$th derivative but cant.

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See here. –  Mhenni Benghorbal Apr 14 '14 at 4:37
If you are interested at the derivative at a fixed point (namely, $0$), you can look at its series expansion. –  Clement C. Apr 14 '14 at 4:37
@MhenniBenghorbal Most of the answers in the linked thread rely critically on the partial fraction expansion, which won't be directly relevant here (unless the OP knows a bit of complex arithmetic). –  user61527 Apr 14 '14 at 4:39

$$\frac1{1+x^2}=\frac1{(x+i)(x-i)}=\frac1{2i}\left(\frac1{x-i}-\frac1{x+i}\right)$$
Use $$\frac{d^n\{(x+a)^m\}}{dx^n}=m(m-1)\cdots(m-r+1)(x+a)^{m-n}$$
Note that this function you have here is the derivative of arctan(x). The Taylor's series of arctan(x) is $x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^9}{9}+\frac{x^{11}}{11}-\frac{x^{13}}{13}...$ Integrate that Taylor series with C=1 (because you have $\frac{1}{1+0^2}=1$) then you should have the Taylor's series for $\frac{1}{1+x^2}$. Then you can move on from there :)