Let $f :(-1,1) \to \mathbb{R};\ \ f(x)=\sum\_{n=0}^\infty a\_n x^n$ be an analytical function expressible as a power series.


Also, let 


$$g : (-1,1) \to \mathbb{R}; \ \ g(x)=\frac{d}{dx} \log{f(x)} = \frac{\sum\_{n=1}^\infty n a\_n x^{n-1}}{\sum\_{n=0}^\infty a\_n x^n} =\sum\_{n=0}^\infty d\_n x^n$$


Assume that $\log f(x)$ is defined for all $\lvert x\rvert<1$.
 Is it possible to obtain a non-recursive expression for the coefficients $d\_n$?


I came across some other answers related to quotients of power series (like [this one](https://mathoverflow.net/questions/53384/power-series-of-the-reciprocal-does-a-recursive-formula-exist-for-the-coeffic "this one") or [this other one](https://math.stackexchange.com/questions/2050216/dividing-an-infinite-power-series-by-another-infinite-power-series# "this other")), but they all rely on recursive expressions. I wonder whether there is a closed form for expressing those coefficients.



Non-recursive expression for coefficients of the derivative of the logarithm of a power series
You start with Faà de Bruno formula (iterated chain rule):
We have for $I, J$ open subsets of $\mathbb R$, $f: I\rightarrow J$, $g: J\rightarrow \mathbb R$, smooth functions, $k\in \mathbb N^{\*}$, 
$$
\frac{(g\circ f)^{(k)}}{k!}=\sum\_{1\le r\le k}\frac{g^{(r)}\circ f}{r!}
\sum\_{\substack{(k\_{1},\dots, k\_{r})\in {(\mathbb N^{\*})}^{r}\\k\_1+\dots+k\_r=k}}\prod\_{1\le j\le r}\frac{f^{(k\_{j})}}{k\_{j}!}.
\tag{$\ast$}$$
It is enough to take $g=\log$, note that for $r\ge 1$, $g^{(r)}(y)=(-1)^{r-1}(r-1)!y^{-r}$, so that you get (assuming $f(0)\not=0$ or something giving sense to the composition)
$$
d\_k=\sum\_{1\le r\le k}
(-1)^{r-1}\frac1rf(0)^{-r}
\sum\_{\substack{(k\_{1},\dots, k\_{r})\in {(\mathbb N^{\*})}^{r}\\k\_1+\dots+k\_r=k}}\prod\_{1\le j\le r}a\_{k\_j}.
$$


