Tìm đạo hàm cấp cao ${y^{(10)}}$ của hàm số $y = {x^2}{e^{2x}}$

Tìm ${{y}^{(10)}}$ của $y={{x}^{2}}{{e}^{2x}}$
Đặt$u={{x}^{2}}$
Đặt $v={{e}^{2x}}$
Ta có ${{y}^{(n)}}=\sum\limits_{k}^{n}{C_{n}^{k}}{{u}^{(n)}}{{v}^{(n-k)}}$
Lại có
$C_{n}^{0}=\frac{n!}{n!.0!}=1$
$C_{n}^{1}=\frac{n!}{(n-1)!.1!}=n$
$C_{n}^{2}=\frac{n!}{(n-2)!.2!}=\frac{1}{2}n.(n-1)$
$C_{n}^{3}=\frac{n!}{(n-3)!.3!}=\frac{1}{6}n.(n-1)(n-2)$
Ta có
$$\begin{align}
& u={{x}^{2}}\Rightarrow {{u}^{(0)}}={{x}^{2}} \\
& {{u}^{(1)}}=2x \\
& {{u}^{(2)}}=2 \\
& {{u}^{(3)}}=0 \\
\end{align}$$
Ta có
$$\begin{align}
& v={{e}^{2x}}=>{{v}^{(n)}}={{2}^{n}}{{e}^{2x}} \\
& {{v}^{(n-1)}}={{2}^{n-1}}{{e}^{2x}} \\
& {{v}^{(n-2)}}={{2}^{n-2}}{{e}^{2x}} \\
\end{align}$$
$$\begin{align}
& \Rightarrow {{y}^{(10)}}={{x}^{2}}{{2}^{10}}{{e}^{2x}}+10.2x{{.2}^{9}}.{{e}^{2x}}+\frac{1}{2}{{.10.9.2.2}^{8}}.{{e}^{2x}} \\
& ={{x}^{2}}{{2}^{10}}{{e}^{2x}}+{{10.2}^{10}}.x.{{e}^{2x}}+{{90.2}^{8}}{{e}^{2x}} \\
& ={{x}^{2}}{{2}^{10}}{{e}^{2x}}+{{5.2}^{11}}.x.{{e}^{2x}}+{{45.2}^{9}}.{{e}^{2x}} \\
\end{align}$$