Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
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?
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.
By induction, you show that if we set (presuming $a$ and $b$ are not zero)
$$f(x) = \ln(ax+b),$$ then we have $$f^n(x)=\frac{a^n(-1)^{n-1}(n-1)!}{(ax+b)^n}$$
