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 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.
