Find the general expression for the n-th derivative I want to find the $n$-th derivatives of the function $$f(x)=\frac{1}{(1-x^2)^b}$$  with respect to x. Here $b$ is a positive constant. By using the chain rule, I can get
the first derivative is $$f'(x)=(-b)(1-x^2)^{-b-1}(-2x).$$ By using the chain rule and product rule, I can get
the second derivative which is $$f''(x)=(-b)(-b-1)(1-x^2)^{-b-2}(-2x)^2+(-b)(1-x^2)^{-b-1}(-2).$$ Again, the third derivative is  $$f'''(x)=(-b)(-b-1)(-b-2)(1-x^2)^{-b-3}(-2x)^3+(-b)(-b-1)(1-x^2)^{-b-2}2(-2x)(-2)+(-b)(-b-1)(1-x^2)^{-b-2}(-2x)(-2).$$ The fouth derivative is  $$f^{(4)}(x)=(-b)(-b-1)(-b-2)(-b-3)(1-x^2)^{-b-4}(-2x)^4+(-b)(-b-1)(-b-2)(1-x^2)^{-b-3}3(-2x)^2(-2)+(-b)(-b-1)(-b-2)(1-x^2)^{-b-3}2(-2x)^2(-2)+(-b)(-b-1)(1-x^2)^{-b-2}2(-2x)(-2)+(-b)(-b-1)(-b-2)(1-x^2)^{-b-3}(-2x)^2(-2)+(-b)(-b-1)(1-x^2)^{-b-2}(-2)(-2).$$
My question is: Is there a general formula or a pattern for the $n$th derivative of $f(x)$? Any suggestions or comments would be very welcome. Thanks in advance.
 A: I suggest you write $$f(x)=(1-x)^{-b}(1+x)^{-b}$$ When you take the derivative of $(1-x)^{-b}$ $n$ times you get $$(-1)^n(1-x)^{-b-n}(-b)(-b-1)...(-b-n+1)=(1-x)^{-b-n}\frac{(b+n-1)!}{(b-1)!}$$ Similarly, you can write an expression for taking the $m$-th derivative of $(1+x)^{-b}$.
Now all you need to take the derivative of order $N$ is to sum all elements with derivative of order $n$ of the first term multiplied with derivative of order $m$ of the second term, using the constraint $N=n+m$
A: I computed the $n^{th}$ derivative with Mathematica. Not sure how you would derive it but since you just wanted the formula here it is.
\begin{align*}
f(x)&=\frac{1}{(1-x^2)^b}\\ f^{(n)}(x)&=n!(1-x^2)^{-b}(x+1)^{-n}\binom{-b}{n} \, _ 2F_1 \left(b,-n;-b-n+1;\textstyle{\frac{x+1}{x-1}} \right)
\end{align*}
Over here $\,_2F_1(a,b;c;d)$ denotes the Ordinary Hypergeometric function.
This closed form matches with the examples you provided.
A: If your objective is to automatize and check the computations you have already done, an idea is to consider:
$$f(x)=\underbrace{(1-x^2)^{-b}}_{\phi(\psi(x))} \ \text{with} \ \phi(x)=x^{-b} \ \text{and} \ \psi(x)=1-x^2$$
and then use the Faa di Bruno formula for the $n$th derivative of the composition of two functions, as given for example in this answer.
Remark: this formula allows to obtain directly $f^{(n)}$ the $n$ without needing to compute the first $(n-1)$ derivatives.
