$\frac{\mathrm d^{22}}{\mathrm dx^{22}}arctg(x^2)$ at $x=0$ Problem:
$$\frac{\mathrm d^{22}}{\mathrm dx^{22}}arctg(x^2)$$ at $x=0$.
My attempt:
$$f'(x)=\frac{2x}{1+x^2}$$
I tried to use General Leibniz rule and I got this.
$$\left(\sum _{k=0}^{22} \binom{22}{k} (2x)^{(k)}*(\frac{1}{1+x^2})^{(n-k)}\right) $$
I saw that for $k=0,1$ summation make sense because: $(2x)'=2$, $2'=0$
But I had a problem to compute 22 and 21 derivative of$\frac{1}{1+x^2}$.
 A: The power series of $\frac{d}{dx}\arctan x^2$ is
$$
\frac{2x}{1+x^4} = 2x- 2x^5 + 2x^9 -2x^{13} + \cdots.\quad(|x|<1)
$$
Meanwhile, the Taylor series at $x=0$ is given by
$$
f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!} x^n
$$
and so
$$
f'(x)=\sum_{n=1}^{\infty}\frac{f^{(n)}(0)}{(n-1)!}x^{n-1}
$$
Compare coefficient of $x^{21}$, then we get $$-2=\frac{f^{(22)}(0)}{21!}$$.
A: As @symplectomorphic said, it's better to use the Maclaurin series of the arctan function.
In your case it would be
$$arctan(x^2) = \sum_{n=0}^\infty \frac{(-1)^nx^{4n+2}}{2n + 1}$$
All you need to do is differentiate the $x^{22}$ or $n = 5$ term in the series.
BTW, you got $f'(x)$ wrong.
It should be $$f'(x) = \frac{2x}{1 + x^4}$$
A: Let be $\varphi(x)=\arctan(x^2)=f(g(x))$ where $f(x)=\arctan x$ and $g(x)=x^2$.
Use the  Faà di Bruno's formula in the classical form
$$
\varphi^{(n)}(x)=\frac{d^n}{dx^n} f(g(x))
=\sum \frac{n!}{m_1!\,m_2!\,\cdots\,m_n!}\cdot
f^{(m_1+\cdots+m_n)}(g(x))\cdot
\prod_{j=1}^n\left(\frac{g^{(j)}(x)}{j!}\right)^{m_j}
$$
where the sum is over all $n$-tuples of nonnegative integers $(m_1,\ldots,m_n)$ satisfying the constraint
$1\cdot m_1+2\cdot m_2+3\cdot m_3+\cdots+n\cdot m_n=n$ or  the Faà di Bruno's formula expressed in terms of Bell polynomials $B_{n,k}(x_1,\ldots,x_{n−k+1})$
$$\varphi^{(n)}(x)=\frac{d^n}{dx^n} f(g(x))= \sum_{k=1}^n f^{(k)}(g(x))\cdot B_{n,k}\left(g'(x),g''(x),\dots,g^{(n-k+1)}(x)\right).$$
