n'th and (n-1)'th derivative of $\frac{1}{x^n - a}$

Question

I have a function of the form $f(x) = \frac{1}{x^n - a}$, and I need to programmatically find the n'th and (n-1)'th derivative of the function. Since the function has this specific form and that the power is tied to which derivative it is, I was wondering if there's an easier method than simply taking the derivative n times.

Context:
I'm trying to implement the house holder's method as a computer algorithm for root finding $\sqrt[n]{a}$ where I use the highest order of the house holders method since it's the fastest of them use able on that function.

Answer 1

For $f(x) = \frac{1}{x^n - a}$ using the chain rule (for $n > 0$) the first derivative comes out as $f'(x) = \frac{-nx^{n-1}}{(x^n - a)^2} $.

I can't see a way that this will simplify to reduce the complexity of further derivatives, and so I would expect the further derivatives will just get more comlplex and you won't be able to get a simple expression like you are looking for.

Answer 2

I don't know if this will help in any way (and probably it won't) but to determine the $n$-th derivative of $f(x)=(x^n-a)^{-1}$, one could observe that $f(x)=(G\circ F)(x)$, where $F(x)=(x^n-a)$ and $G(x)=x^{-1}$. Then one could apply Faa di Bruno's formula

$$(F\circ G)^{(n)}(x) = \sum \frac{n!}{b_1!\cdots b_n!}G^{(r)}(F(x))\prod_{i=1}^n \left(\frac{F^{(i)}(x)}{i!}\right)^{b_i}$$

where the sum is over all integer partitions of $n$, $r=b_1+\cdots+b_n$, and observe that $G^{(r)}(y)=(-1)^{r}r!y^{-(r+1)}$, and a simpler expression for $F^{(i)}$ also arises (e.g. $F^{(n)}=n!$).