Find $f^{(100)}(0) $ and $f^{(101)}(0) $ if $f(x)=xe^{\arctan{x}}$ $$f(x)=xe^{\arctan{x}}$$
Part of my solution
$$f^{(n)}(x)=\sum_{k=0}^{n}\binom{n}{k}x^{(k)}(e^{\arctan{x}})^{(n-k)}=x(e^{\arctan{x}})^{(n)}+(e^{\arctan{x}})^{(n-1)}$$
First term probably disappears because $x=0$ but i don't know what to do with second term.  

Partial answer:
As you pointed out,
$$\begin{align}f^{[n]}(x)&=\sum_{i=0}^n{n\choose i}x^{[i]}\left(e^{\tan^{-1}(x)}\right)^{[n-i]}
\\
&=x\left(e^{\tan^{-1}(x)}\right)^{[n]}+n\left(e^{\tan^{-1}(x)}\right)^{[n-1]}
\end{align}$$
We set $g(x)=\left(e^{\tan^{-1}(x)}\right)$, $h(x)=\tan^{-1}(x)$ and consider $g^{[n]}(x)$.
$\begin{align}g'(x)&=\left(e^{h(x)}\right)'
\\
&=h'(x)g(x)
\\
g''&=h''g+h'g'
\\
&=h''g+h'h'g
\\
&=(h''+(h')^2)g
\\
g'''&=(h'''+2h'h'')g+(h''+(h')^2)h'g
\\
&=(h'''+3h'h''+(h')^3)g
\\
g^{[4]}&=(h^{[4]}+3(h''h''+h'h''')+3(h')^2h'')g+(h'''h'+3(h')^2h''+(h')^4)g
\\
&=(h^{[4]}+4(h')^2h''+3(h'')^2+4h'h'''+(h')^4)g
\\
g^{[5]}&=(h^{[5]}+4(2(h')h''h''+(h')^2h''')+6(h'')h'''+4(h''h'''+h'h^{[4]})+4(h')^3h'')g\dots
\\
&\quad\dots+(h'h^{[4]}+4(h')^3h''+3h'(h'')^2+4(h')^2h'''+(h')^5)g
\\
&=(h^{[5]}+8(h')^3h''+11h'(h'')^2+8(h')^2h'''+10h''h'''+5h'h^{[4]}+(h')^5)g
\end{align}$
Note that each bracketed term is a sum of the form:
$$\sum_{i=0}^{n}C_i\prod\left(h^{[m]}\right)^p$$
Where $C_i$ are integer constants and for each $i$, $\sum{pm}=n$. In fact, all sums of $pm$ equal to $n$ are represented as terms. For example, $3=(3\cdot1)=(1+2)=(1\cdot3)$ so there are terms with $(h')^3$, $h'h''$ and $h'''$. This is strikingly similar to the binomial series but I've yet to determine the constants. We would probably be able to use Faa di Bruno's formula for this.
We know that $h'(x)=\frac{1}{x^2+1}$, which we equate to a geometric series in $-x^2$ and generalise.
$$\begin{align}
h'(x)&=\sum_{i=0}^\infty(-x^2)^i
\\
h''(x)&=\sum_{i=1}^\infty(-1)^i\cdot2ix^{2i-1}
\\
h^{[n]}(x)&=\sum_{i=n-1}^\infty(-1)^i(2i)_{n-1}x^{2i-(n-1)}
\end{align}$$
Where $(x)_n=\prod_{i=0}^{n-1}(x-i)$ and $x^0=1$ when $x=0$.
Hence, $h'(0)=1, h''(0)=0, h'''(0)=-2, h^{[5]}=4!,h^{[7]}=-6!\dots$ and in general:
$$h^{[n]}(0)=\begin{cases}0&n\:\mathrm{even}
\\
(-1)^{\frac{n-1}{2}}(n-1)!&n\:\mathrm{odd}
\end{cases}$$
We may substitute this into the multi-chain-rule expansion of $g^{[n]}$(x) at $x=0$ using that $g(0)=1$.
$\begin{align}g'(0)&=1
\\
g''(0)&=1
\\
g'''(0)&=-2+1
\\
g^{[4]}(0)&=4(-2)+1
\\
g^{[5]}(0)&=4!+8(-2)+1
\dots
\end{align}$
The problem becomes finding the general coefficients $C_i$ of the odd terms in the expansion of $g^{[n]}$ but I'm unsure how to proceed.
