Finding the $n$th derivative of $1/(1-a(1-x)(1+x))$ with respect to $x$.

Question

I need to find the $n$th derivative of $$\frac{1}{1-a(1-x)(1+x)}$$ with respect to $x$. Wolfram says that when $a=1$ we have the $n$th derivative as $$\frac{1}{1-(1-x)(1+x)}^{(n)} = (-1)^n \frac{(n+1)!}{x^{n+2}}$$ however I'm unable to use this information to find a closed form. I tried to use some other $a$ but wolfram fails at that, maybe because I'm using the free version and I don't have bought the full version. Any help is appreciated, thanks.

Answer 1

Assume $a\ne1$ and let $b\in\mathbb C$ be a square root of $\frac a{a-1}.$

$$\frac1{1-a(1-x)(1+x)}=\frac1{1-a}\frac1{1-(bx)^2}=\frac1{2(1-a)}\left(\frac1{1-bx}+\frac1{1+bx}\right).$$ The $n$-th derivative should now be easy to compute.