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.

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.
