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Is there a closed-form solution for $\frac {d^n}{dx^n}\frac 1{f(x)}$? I've looked at the first five derivatives in search of some pattern, but I can't identify anything strong enough to give a closed formula.

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  • $\begingroup$ Is $f(x)$ a particular function or its just a generalization? $\endgroup$ – AvZ Mar 6 '15 at 12:45
  • $\begingroup$ @AvZ A generalization. $\endgroup$ – Laertes Mar 6 '15 at 13:01
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It's not exactly a closed form, but Faà di Bruno's formula

$$\frac{d^n}{dx^n} f(g(x))= \sum_{k_1 + 2k_2 + \ldots + n k_n = n} \frac{n!}{k_1!1!^{k_1}...k_n!n!^{k_n}}\left(\frac{d^{k_1 + \ldots+ k_n}}{dx^{k_1 + \ldots+ k_n}} f\right)(g(x)) \prod_j \left(\frac{d^j}{dx^j} g(x) \right)^{k_j}$$

gives

$$\frac{d^n}{dx^n} \frac{1}{g(x)} = \sum_{k_1 + 2k_2 + \ldots + n k_n = n} \frac{(-1)^k n!({k_1 + \ldots+ k_n})!}{k_1!1!^{k_1}...k_n!n!^{k_n}g(x)^{1+k_1 + \ldots+ k_n}} \prod_j \left({\frac{d^j}{dx^j} g(x)} \right)^{k_j}$$

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  • $\begingroup$ Appreciated, but I was hoping for something slightly more closed. $\endgroup$ – Laertes Mar 6 '15 at 13:31
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    $\begingroup$ @Laertes Aside from perhaps a recurrence relationship, I think this is the best you can hope for I'm afraid! $\endgroup$ – Thomas Russell Mar 6 '15 at 13:53
  • $\begingroup$ Small correction: The $(-1)^k$ in the second formula makes no sense because $k$ is not defined. It should read $(-1)^{k_1+\dots+k_n}$. $\endgroup$ – WillG Oct 18 '19 at 0:37

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