Simplified expression for the nth derivative of the n+1th power of a function

Question

$$\frac{d^n}{dx^n}(f(x))^n \text{ and similar expressions}$$

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I was trying to work out the Taylor series for the solution to the general cubic using Lagrange Inversion (out of curiosity: I was wondering which of the roots the series would give for different cubics) and got stuck on finding the $n^{th}$ derivative of $ (\frac{1}{ax^2+bx+c})^{n+1} $.

First I tried working it out by hand but I couldn't see any obvious pattern that would allow me to compute the general case. Wolfram Alpha was able to calculate the next few values of n and I noticed some things but there was no obvious pattern for the general term.

I realized I could do it by employing the Laplace Transform (or the Fourier Transform or maybe even directly from the Taylor Series) where taking n+1 derivatives is trivial but found it rather difficult to compute any of these in general since the powers of $ \frac{1}{ax^2+bx+c}$ are rather chaotic. I am a bit stuck on this.

Also, if someone knows how to do this, I'm now curious if there is a way to get the general ${n-1}^{th}$, $n^{th}$, or ${n+1}^{th}$ derivatives of the $n^{th}$ or ${g(n)}^{th}$ powers of a function f(x) for some function g(n), and as to where the coefficients of these terms come from (some sort of analogue of Pascal's or one of Stirling's triangles maybe?).

For example, the coefficients of the $n^{th}$ derivatives of $(f(x))^n$ can be arranged in a triangle like this when you arrange them in increasing order in each row:

$$\text{1}$$

$$\text{2 2}$$

$$\text{3 6 18}$$

$$\text{4 24 36 48 144}$$

But I don't remember seeing these numbers anywhere and couldn't even find this sequence (read out by rows) in the OEIS.

Answer 1

There is the broader generalized product rule which states $$ (f_1 \cdot \ldots\cdot f_m)^{(n)} = \sum_{k_1+k_2+\cdots+k_m=n} {n \choose k_1, k_2, \ldots, k_m} \prod_{1\le t\le m}f_{t}^{(k_{t})} $$ Take $m=n$ and $f_i = f$ for all $i$ to get $$ \frac{d^n f^n}{dx^n} = \sum_{k_1 + \cdots + k_n = n} {n \choose k_1, k_2, \ldots, k_n} \prod_{j=1}^n \frac{d^{k_j} f}{dx} $$ These coefficients form the multinomial coefficients, a generalization of the binomial coefficients, which arise from the expansion of $(x+y)^n$, whereas $$ \binom{n}{k_1,\cdots,k_m} := \frac{n!}{k_1! \cdots k_m!} $$ arises from the expansion of $(x_1+\cdots+x_m)^n$.

One can tackle the order $n \pm 1$ cases in the obvious way from here. And I guess order $g(n)$ but I imagine the expression there would range on the nigh useless considering this barely feels adequate. But maybe someone knows something more clever.