What is the $m$th derivative of $\log\left(1+\sum\limits_{k=1}^N n_kx^k\right)$ at $x=0$? Let $n_k$ be integers. Is there a general formula for the Taylor expansion of
$\log(1+\sum_{k=1}^N n_kx^k)$ at $x=0$?
This boils down to find an expression for the $m$th derivative of  $\log(1+\sum_{k=1}^N n_kx^k)$ evaluated $x=0$:
$$
\frac{d^m}{dx^m}\log(1+\sum_{k=1}^N n_kx^k)\Biggr|_{x=0}?
$$
Expanding a (work-in-progress) example like $-\log(n_3x^3+n_2x^2+n_1x+1)$ gives:
$$
\begin{array}{cl}
 x&            (- n_1)\\ 
+ \frac{x^2}2& (+ n_1^2 - 2n_2 ) \\
+ \frac{x^3}3& (- n_1^3 + 3n_2 n_1   - 3n_3) \\
+ \frac{x^4}4& (+ n_1^4 - 4n_2 n_1^2 + 4n_3 n_1  + 2n_2^2 ) \\
+ \frac{x^5}5& (- n_1^5 + 5n_2 n_1^3 - 5n_3n_1^2 - 5n_2^2 n_1  + 5n_3n_2    ) \\
+ \frac{x^6}6& (+ n_1^6 - 6n_2 n_1^4 + 6n_3n_1^3 + 9n_2^2n_1^2 - 12n_3n_2n_1 - 2n_2^3 +3 n_3^2  ) \\
&\dots
\end{array}
$$
Some findings:


*

*For each $\frac{x^k}k$ the factor in brackets relates to partitions of $k$, which is not very surprising. 

*Fixing $k$ then the sign is a term inside a bracket can be determined by the sum of powers of $n_j$: Even sums are negative (e.g. $-n_1^4$) and odd sums positive (e.g. $+5n_2^2n_1$), which reminds me on Möbius' function...

*From one line to the other you can see that multiplying a $n_1$ to the second term adds $1$ to the prefactor (and changes sign)
This looks like a combinatorial thing...
 A: Slightly different from Peter's answer. From $$f(x)=\log{\left(1+\sum_{k=1}^Nn_kx^k\right)}=\log{g(x)} \Rightarrow f'(x)=\frac{g'(x)}{g(x)} \Rightarrow f'g=g'$$ 
With the latter, applying  General Leibniz rule and considering $g(0)=1$, a pattern emerges and binomial coefficients. For example:
Case $m=1$
$$f'(x)=\frac{g'(x)}{g(x)} \Rightarrow f'(0)=\frac{g'(0)}{g(0)}=n_1$$
Case $m=2$
$$f'g=g' \Rightarrow f''g+f'g'=g'' \Rightarrow f''(0)+n_1^2=2n_2$$
$$f''(0)=2n_2-n_1^2$$
Case $m=3$
$$f''g+f'g'=g'' \Rightarrow f^{(3)}g+2f''g'+f'g''=g^{(3)} \Rightarrow f^{(3)}(0)+2(2n_2-n_1^2)n_1+2n_1n_2=3\cdot2n_3$$
$$f^{(3)}(0)=2n_1^3-6n_1n_2+6n_3$$
Case $m=4$
$$f^{(3)}g+2f''g'+f'g''=g^{(3)} \Rightarrow f^{(4)}g+3f^{(3)}g'+3f''g''+f'g^{(3)}=g^{(4)}$$
and so on. Now we can define two sequences
$$g_n=g^{(n)}(0),g_0=g(0)=1,g_1=g'(0)=n_1$$
$$f_n=f^{(n)}(0),f_0=f(0)=0,f_1=f'(0)=n_1$$
$$g_{n+1}=\sum_{k=0}^n\left(\binom{n}{k} f_{n-k+1}g_{k}\right)=f_{n+1}g_{0}+\sum_{k=1}^n\left(\binom{n}{k} f_{n-k+1}g_{k}\right)$$
or 
$$f_{n+1}=g_{n+1}-\sum_{k=1}^n\left(\binom{n}{k} f_{n-k+1}g_{k}\right)=g_{n+1}-n\cdot n_1\cdot f_n-n_1\cdot g_n-\sum_{k=2}^{n-1}\left(\binom{n}{k} f_{n-k+1}g_{k}\right)$$
Result is, there is no need to compute derivatives of $f(x)$ at $x=0$ directly and $g(x)$ is a polynomial whose derivatives are easier to calculate.
A: Using Faà di Bruno's formula the $n$th derivative is expressible as a sum of bell polynomials. In sympy for your example:
from sympy import *

x,n1,n2,n3=symbols('x,n1,n2,n3')

def f(x):
    return -log(x)
def g(x):
    return n3*x**3+n2*x**2+n1*x+1

n = 3;

print
print diff(f(g(x)),x,n).subs(x,0)

v = map(lambda k: diff(g(x),x,k).subs(x,0), range(1,n+1))
s = 0
for k in range(1,n+1):
    s +=diff(f(x),x,k).subs(x,g(0))*bell(n,k,v)
print s

In this case
diff(g(x),x,k)

evaluates to $n_1 + 2n_2x + 3n_3x^2$ for $k=1$, $2n_2 + 6n_3x$ for $k=2$, $6n_3$ for $k=3$ and $0$ for $k\geq4$. Setting $x=0$ we get $(g'(0),g''(0),\dots,g^{(n-k+1)}(0))=(n_1,2n_2,6n_3,0,\dots,0)$. Replace the code for $v$ with
 v=n*[0];
 v[0]=n1;v[1]=2*n2;v[2]=6*n3;

Furthermore since $f'(x)=-1/x$, $f''(x)=+1/x^2$, $f'''(x)=-2/x^3\dots$ and $g(0)=1$ the line
 s +=diff(f(x),x,k).subs(x,g(0))*bell(n,k,v)

is equivalent to 
 s +=(-1)^k*factorial(k-1)*bell(n,k,v)

A: The following is in terms of formal power series. I don't bother about convergence radii.
Assume that the functions $f(x)=\sum_{l\geq 0} f_l x^l$ and $q(x)=\sum_{k\geq0}q_kx^k$, $q_0=1$,  are related by
$$f(x)=\log q(x)\ .\tag{1}$$
Then $f_0=0$, and $(1)$ implies $$q(x)f(x)=q'(x)\ ,$$
or unpacked:
$$\sum_{k\geq0}q_kx^k\cdot\sum_{l\geq 1}l f_l x^{l-1}=\sum_{r\geq1}r q_r x^{r-1}\ .$$
This can be written as
$$\sum_{r\geq0}\left(\sum_{k+l=r} q_k(l+1)f_{l+1}\right)x^r=\sum_{r\geq0}(r+1)q_{r+1} x^r\ ,$$
and comparing coefficients leads to
$$\sum_{k+l=r}q_k(l+1)f_{l+1}=(r+1)q_{r+1}\qquad(r\geq0)\ .$$We now take the summand $k=0$, $l=r$ out of the sum on the left hand side and so obtain the following recursion for the $f_l$:
$$f_{r+1}=q_{r+1}-\sum_{l=1}^r{l\over r+1} q_{r+1-l}\>f_l\qquad(r\geq0)\ .\tag{2}$$
A: You've rediscovered the classical Faber polynomials, presented in OEIS A263916.
