Hello, I'm trying to find the $n$th derivative of a function, where $n\in\mathbb N$ $$\frac{d^{n-1}}{dx^{n-1}} \!\!\left[f(x)^n\right]$$ I'm looking for some kind of sum or product (or nesting of these) which gives the value above.
So I tried evaluating the function at some values of $n$
$$ \begin{array}{lr} f(x) & \text{for } n=1\\ 2f(x)f^{(1)}(x) & n=2\\ 3f(x)^2f^{(2)}(x)+6f(x)f^{(1)}(x)^2&n=3\\ 4f(x)f^{(4)}+24f(x)f^{(1)}(x)+\cdots\\ \qquad36f(x)^2f^{(1)}(x)f^{(2)}(x) & n=4 \end{array} $$
Some things I noticed was
1. In expanded form, it always had a term of the form
$$nf(x)^{n-1}f^{(n-1)}(x)$$
2. In expanded form you could always factor out $nf(x)$ from all terms
3. $f$ is always multiplied together with itself (or a derivative of itself) $n$ times in each term.
4. There's always $n-1$ terms, in the case of $n=1$ it seems to be the always being there factor of $nf(x)$ multiplied with the empty product.
Tip: I have found the wolfram alpha input D[f[x]^n, {x,n-1}], n -> (insert n here) very useful for calculating the $n$'th version of this formula.
Note: The notation below means the $m$th derivative of $f$
$$f^{(m)}(x)$$
There is a formula for
$$\frac{d^n}{dx^n}\left(f(g(x))\right) $$
(this suits you). This is called Faà di Bruno's formula https://en.wikipedia.org/wiki/Fa%C3%A0_di_Bruno%27s_formula, use it properly and you will get a result