TITLE: Find the nth derivative
QUESTION [7 upvotes]: Find the $n$th derivative of the function 
  $$y=\ln(ax+b).$$

I have computed the following derivatives:
$$y'=\frac{a}{ax+b}$$
$$y''=\frac{-a^2}{(ax+b)^2}$$
$$y'''=\frac{2a^3}{(ax+b)^3}$$
I think
$$y^{(n)}=\frac{(-1)^n c a^n}{(ax+b)^n}$$
But I could not determine the pattern for the constant $c$
How can I determine it?

REPLY [10 votes]: By induction, you show that if we set (presuming $a$ and $b$ are not zero)
$$f(x) = \ln(ax+b),$$
then we have
$$f^n(x)=\frac{a^n(-1)^{n-1}(n-1)!}{(ax+b)^n}$$

REPLY [10 votes]: We can show by induction $y^{(n)}=\frac{(-1)^n(n-1)!a^n}{(ax+b)^n}$
For the inductive step note that we can write the $n+1$-th derivative as:
$\frac{d}{dx} (-1)^n(n-1)!a^n(ax+b)^{-n}=(-1)^n(-n)(n-1)! a^{n+1}(ax+b)^{-n-1}=\frac{(-1)^{n+1}n!a^{n+1}}{(ax+b)^{n+1}}$
The second equality follows from the chain and power rules.