I need to compute higher order derivatives (7th 8th, 9th, 10th) evaluated at $x_0 = 0$ of function defined as
$$f(x) = \begin{cases} \frac1{8!} & x=0 \\ \frac{\log(x+1)}{8!x} & x\in(-1,\infty) \setminus \{0\}\end{cases}$$
What worked in similar problems was writing Taylor expansion of $g(x) = \log(x+1)$ at $0$ and then shifting it by one power of $x$, like: $$\log(x+1) = \sum_{k=0}\frac{(x-0)^k}{k!}g^{(k)}(0),$$
where $g(x) = \log(x+1)$ and $g^{(k)}(0) = (-1)^{k+1}(k-1)!(x+1)^{-k}.$ Then $$\frac{\log(x+1)}{8!x} = \sum_{k=0}\frac{x^k}{k!}(-1)^{(k+1)}(k-1)!\frac{1}{8!x} = \\\sum_{k=0}\frac{x^{k-1}}{(k-1)!}\cdot\frac{(-1)^{k+1}}{k\cdot8!}(k-1)!$$
Then given $t = k-1$ left part under sum would be $\frac{(x-0)^t}{t!}$ and right would be $f^{(t)}(0)$. But doing so we should change $\sum_{k=0}$ to $\sum_{t = -1}$ - it doesn't match Taylor expansion formula (summation should start at $0$ and I don't know what to do about that fact.
The right answer to the problem is: $f^{(7)}(0) = -1/64$, $f^{(8)}(0) = 1/9$, $f^{(9)}(0) = -9/10$, $f^{(10)}(0) = 90/11$.
$$\sum_{n=1}^\infty \frac{(-1)^{n+1}x^n}n=\ln(x+1)$$
Therefore:
$$\sum_{n=1}^\infty\frac{(-1)^{n+1}x^{n-1}}n=\frac{\ln(x+1)}x$$
or
$$f(x)=\sum_{n=1}^\infty\frac{(-x)^{n-1}}{8!n}= \sum_{n=0}^\infty\frac{(-x)^nn!}{8!(n+1)n!} =\frac{\ln(x+1)}{8!x}$$
in this Interactive graph and $f^{(n)}(0)=\frac{(-1)^nn!}{8!(n+1)}$ with derivatives from the Taylor series formula matching yours