$2n$ th partial derivative of $\frac{1}{y(1+x^2)-1}$ with respect to $x$.

Question

I need to find the $2n$ th derivative with respect to $x$ of the function $f = \frac{1}{y(1+x^2)-1}$. I tried differentiating util a pattern was founded, but that didn't happen.

I think the $x^2$ is making everything more difficult so I wanted to know if is possible to write $f$ as a sum of functions that don't depend on $x^2$. Also, I'm talking about a closed form.

To make it more clear of what I'm talking about I will give a example. To find the $n$ th derivative of $1/(1-x^2)$ I tried differentiating the function to find a pattern but didn't work. But if we write $1/(1-x^2) = 1/(2x+2) - 1/(2x-2) $ we can differentiate the first and second term separately and easily find a pattern then just add the two together.

Answer 1

As Jimmy Sabater commented, consider that $$y(1+x^2)-1=y(x-r_1)(x-r_2) \qquad \text{where}\qquad r_{1,2}=\pm\frac{\sqrt{1-y}}{\sqrt{y}}$$ Now, use partial fraction decomposition $$\frac 1 {y(1+x^2)-1}=\frac 1{y(r_1-r_2)}\left(\frac 1 {x-r_1}-\frac 1 {x-r_2} \right)$$ and any derivative becomes quite simple.