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?

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    $\begingroup$ For the fourth derivative, $c = 6$. For the fifth, we have $c = 24.$ Using induction, you can show that $c = (n-1)!$ in general. $\endgroup$
    – user328442
    Dec 31, 2017 at 22:29
  • 1
    $\begingroup$ Hey ! if your question has been answered, please consider accepting an answer :) $\endgroup$ Dec 13, 2019 at 12:08

2 Answers 2


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 $$f^n(x)=\frac{a^n(-1)^{n-1}(n-1)!}{(ax+b)^n}$$


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