# 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.

• Well, there is always the general Leibniz Rule, Faa di Bruno, etc, but whether or not they will generate pretty results in this particular case is doubtful. Sep 21 '21 at 21:04
• Perhaps you can use the following idea. We have $\frac{1}{1-x^2}=\frac{1}{1-x}+\frac{1}{1+x}$. So $$\frac{1}{(1-x^2)^b} = \left( \frac{1}{1-x} + \frac{1}{1+x} \right)^b$$ Sep 22 '21 at 1:58

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 $$N=n+m$$

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.

• [+1] Interesting, I wouldn't have guessed such a closed for expression exist. My own answer is somewhat more "classical". Sep 21 '21 at 17:42
• Does the hypergeometric function simplify to something nice for $b$ (and $n$) integer?
– Joe
Sep 22 '21 at 5:59
• I just tried that, assuming b, n>0 and b, n $\in \mathbb{Z}$ the Hypergeometric term doesn't seem to simplify. When b and n are integers however the Pochammer terms in the Hypergeometric series can be written in terms of a bunch of Gamma functions. I haven't had any luck in trying to simplify it from there however. Sep 22 '21 at 6:51

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.

• An interesting article about Faa di Bruno formula:maa.org/sites/default/files/pdf/upload_library/22/Ford/… Sep 21 '21 at 18:04
• Could you explain the remark a bit? Looking at both links I don't see how you can avoid computing all the lower derivatives Sep 22 '21 at 5:35
• @perpetuallyconfused I meant computing $f^{(n)}$ without its previous derivatives. Of course we need the different derivatives of $\phi$ and $psi$. Sep 22 '21 at 6:04