Find y^((n))(0) for y = ln (ax+b), b>0
Find $y^{(n)}(0)$ for $y = \ln\bigl(ax+b\bigr)$, $b>0$.
Explanation of Method
The logarithm of Problem 288, with a linear argument.
Step 1: The first few derivatives
The chain rule contributes a factor $a$ at every step: \[y' = \frac{a}{ax+b}, \qquad y'' = -\frac{a^{2}}{\bigl(ax+b\bigr)^{2}}, \qquad y''' = \frac{2a^{3}}{\bigl(ax+b\bigr)^{3}}.\]
Step 2: The general formula
\[ y^{(k)} = \frac{\bigl(-1\bigr)^{k-1}\bigl(k-1\bigr)!\,a^{k}}{\bigl(ax+b\bigr)^{k}}, \qquad k\ge1. \tag{1} \]
Induction. The case $k = 1$ is the first line. Assuming (1) and differentiating:
\[ y^{(k+1)} = \bigl(-1\bigr)^{k-1}\bigl(k-1\bigr)!\,a^{k}\cdot\frac{-k\,a}{\bigl(ax+b\bigr)^{k+1}} = \frac{\bigl(-1\bigr)^{k}k!\,a^{k+1}}{\bigl(ax+b\bigr)^{k+1}}, \]
which is (1) with $k+1$ in place of $k$.
Step 3: Evaluate at the origin
At $x = 0$ the denominator is $b^{k}$, and $b>0$ guarantees this is legitimate: $y^{(n)}(0) = \frac{\bigl(-1\bigr)^{n-1}\bigl(n-1\bigr)!\,a^{n}}{b^{n}}$.
Step 4: Verification
\[n = 1:\ y'(0) = \frac{a}{b}; \qquad n = 2:\ y''(0) = -\frac{a^{2}}{b^{2}} \;\;\checkmark\]
\[a = b = 1:\ y^{(n)}(0) = \bigl(-1\bigr)^{n-1}\bigl(n-1\bigr)!, \text{ the case } \ln\bigl(1+x\bigr) \;\;\checkmark\]
\[\text{Series: } \ln\bigl(ax+b\bigr) = \ln b+\sum_{n=1}^{\infty}\frac{\bigl(-1\bigr)^{n-1}}{n}\left(\frac{a}{b}\right)^{n}x^{n} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y^{(n)}(0) = \bigl(-1\bigr)^{n-1}\bigl(n-1\bigr)!\left(\dfrac{a}{b}\right)^{n}, \qquad n\ge1\,}\]
Remark
Remark. Everything depends on $a$ and $b$ only through the ratio $a/b$, which is forced by the identity $\ln(ax+b) = \ln b+\ln\!\bigl(1+\tfrac{a}{b}x\bigr)$: the constant $\ln b$ has no derivatives, and what remains is Problem 288 with $x$ rescaled. The series converges for $\bigl|\tfrac{a}{b}x\bigr|\lt 1$, i.e. as long as $ax+b$ stays positive and bounded away from doubling $b$.