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.

  • $\begingroup$ 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. $\endgroup$
    – K.defaoite
    Sep 21 '21 at 21:04
  • $\begingroup$ Perhaps you can use the following idea. We have $\frac{1}{1-x^2}=\frac{1}{1-x}+\frac{1}{1+x}$. So $$ \frac{1}{(1-x^2)^b} = \left( \frac{1}{1-x} + \frac{1}{1+x} \right)^b $$ $\endgroup$
    – Kapil
    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.

  • 2
    $\begingroup$ [+1] Interesting, I wouldn't have guessed such a closed for expression exist. My own answer is somewhat more "classical". $\endgroup$
    – Jean Marie
    Sep 21 '21 at 17:42
  • 1
    $\begingroup$ Does the hypergeometric function simplify to something nice for $b$ (and $n$) integer? $\endgroup$
    – Joe
    Sep 22 '21 at 5:59
  • $\begingroup$ 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. $\endgroup$ 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.

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

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