Find $9$'th derivative of $\frac{x^3 e^{2x^2}}{(1-x^2)^2}$ How can I find $9$'th derivative at $0$ of $\displaystyle \frac{x^3 e^{2x^2}}{(1-x^2)^2}$. Is there any tricky way to do that?
This exercise comes from discrete mathematic's exam, so I think that tools like taylor  can't be used there.
 A: We know that 
$$\frac{x}{(1-x)^2}=\sum_{n=1}^{\infty}nx^{n}$$
$x\rightarrow x^2$
$$\frac{x^2}{(1-x^2)^2}=\sum_{n=1}^{\infty}nx^{2n}$$
multiply by $x$
$$\frac{x^3}{(1-x^2)^2}=\sum_{n=1}^{\infty}nx^{2n+1}\tag1$$
$$e^x=\sum_{m=0}^{\infty}\frac{x^m}{m!}$$
$x\rightarrow 2x^2$
$$e^{2x^2}=\sum_{m=0}^{\infty}\frac{2^mx^{2m}}{m!}\tag2$$
so
$$y=\frac{x^3 e^{2x^2}}{(1-x^2)^2}=\sum_{n=1}^{\infty}\sum_{m=0}^{\infty}\frac{n2^mx^{2n+2m+1}}{m!}$$
to get the coefficient of $x^9$ ,we should find the solutions of $2m+2n+1=9$ which are
$$(m,n)=(0,4),(1,3),(2,2),(3,1)$$
so the coefficient of $x^9$ is $\frac{46}{3}$
the 9 'th derivative at $x=0$ will be
$$y(0)^{(9)}=\frac{46}{3}*9!=5564160$$
A: Hint: Take $\ln$ of both sides of
$$y=\frac{x^3 e^{2x^2}}{(1-x^2)^2}$$
and use these facts that 
$$\dfrac{d^n}{dx^n}\ln(1+x)=\dfrac{(-1)^{n-1}(n-1)!}{(1+x)^{n+1}}$$
$$\dfrac{d^n}{dx^n}\ln(1-x)=\dfrac{(n-1)!}{(1-x)^{n+1}}$$
A: A reasonable method here would be to find the Taylor polynomial for each term (i.e. $x^3$, $\frac{1}{(1-x^2)^2}$ and $e^{2x^2}$), each of which is straightforward, and then compute the $x^9$ term.
