What is the $n^\text{th}$ derivative of $f(x)=\frac{1}{1+x^2}$

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?

• your second derivative and your fourth derivative are equal, are you sure that's the case? Jul 8 '18 at 10:44
• @AmateurMathPirate, Thank you. I correct it. Jul 8 '18 at 11:02

$$2f(x)=\frac1{1+ix}+\frac1{1-ix}.$$ Therefore $$2f^{(n)}(x)=\frac{(-i)^nn!}{(1+ix)^{n+1}}+\frac{i^nn!}{(1-ix)^{n+1}}.$$
• Thank you but I want real values and also around a value $a$. Jul 8 '18 at 8:39
• That's fine; whenever $x$ is real, then so is $f^{(n)}(x)$. @user85361 Jul 8 '18 at 8:41
• There isn't any convergence requirement like $\vert x\vert <1$? Jul 8 '18 at 8:43
• @user85361 As you can see, this formula is valid for all real $x$. Jul 8 '18 at 8:44
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.