TITLE: $n$th derivative of $\ln f(x)$
QUESTION [7 upvotes]: I've been trying to find the $n$th derivative of the function $h(x)=\ln f(x)$. Wolfram|Alpha and Gradshteyn's Table of Integrals, Series and Products give 
 $$
    \frac{d^n}{dx^n}\ln f(x)=-\sum_{k=1}^n\frac{(-1)^k\left[\binom{n}{k}\frac{d^n}{dx^n}\left(f(x)^k\right)\right]}{kf(x)^k}.
 $$
Notice, however, that the $\frac{d^n}{dx^n}\left(f(x)^k\right)$ is not explicited. 
Again, Gradshteyn says 
  $$
    \frac{d^n}{dx^n}\left(f(x)^k\right) = 
      k\binom{n-k}{n}\left[\sum_{\ell=0}^n\frac{f(x)^{p-n}}{p-n}\frac{d^n(f(x)^\ell)}{dx^n}\right]
  $$
which is implicit. Knowing that $k\leq n$, is there a way to write the $n$th derivative of $\ln f(x)$ in a way that makes clear the dependance on 
 $$
 \frac{d^m}{dx^m}f(x)\qquad\text{and}\qquad \left(\frac{df(x)}{dx}\right)^\ell?
 $$
I've looked at Leibniz formula for the $n$th derivative of a product with product $f'/f$, the logarithmic derivative and Faà di Bruno's formula, but gotten nowhere.

REPLY [4 votes]: This is a special case of Faà di Bruno's formula:
$$
\frac{d^n}{dx^n} \ln f(x)=\\
\phantom{break}\\
\sum_{m_1+2m_2+\cdots+nm_n=n} \frac{n!}{m_1!\,m_2!\,\,\cdots\,m_n!}
\frac{ (-1)^{m_1+\cdots+m_n-1} (m_1+\cdots+m_n-1)! }{f(x)^{m_1+\cdots+m_n}} \prod_{1\le j\le n}\left(\frac{f^{(j)}(x)}{j!}\right)^{m_j}
$$
The sum is over all $n$-tuples $(m_1,\ldots,m_n)$ such that $m_1, \dots, m_n\ge 0$ and, as indicated in the formula, $\sum_{1\le j\le n} j m_j=n$.