I am attempting to write down a closed from expression for$$\frac{d^n}{dx^n}\log(f(x))$$at $x=x_0$ with the additional assumptions that $f(x_0)=1$ and $f'(x_0)=0$. As mentioned in this post, the computation of $\frac{d^n}{dx^n}\log(f(x))$ is a special case of Faà di Bruno's formula and$$\frac{d^n}{dx^n} \log f(x)=\\\phantom{break}\\\sum_{m_1+2m_2+\cdots+nm_n=n} \frac{n!}{m_1!\,m_2!\,\,\cdots\,m_n!}\frac{ (-1)^{m_1+\cdots+m_n-1} (m_1+\cdots+m_n-1)! }{f(x)^{m_1+\cdots+m_n}} \prod_{1\le j\le n}\left(\frac{f^{(j)}(x)}{j!}\right)^{m_j}.$$Here, 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, $\sum_{1\le j\le n} j m_j=n$.
I was wondering if there exists a further simplified expression for $\frac{d^n}{dx^n}\log(f(x))$ at $x=x_0$ with the simplifying assumptions $f(x_0)=1$ and $f'(x_0)=0$. Further, what would the first few terms of this derivative look like? Summing over tuples is confusing me.