# Prove $\frac{d^n}{dx^n}\ln(x)=\frac{(n-1)!(-1)^{n-1}}{x^n}$ by induction

Prove

$$\frac{d^n}{dx^n}\ln(x)=\frac{(n-1)!(-1)^{n-1}}{x^n}$$

by induction.

## Attempt to solve

Base case

$$n=1$$

$$\frac{d}{dx}\ln(x)=\frac{(1-1)!(-1)^{1-1}}{x^{1}}=\frac{1}{x}$$

which is true.

Induction step

Induction hypothesis: equation is true when $$n=k$$

$$\frac{d^k}{dx^k}\ln(x)=\frac{(k-1)!(-1)^{k-1}}{x^k}$$

Induction conjecture: when $$n=k+1$$

$$\frac{d^{k+1}}{dx^{k+1}} \ln(x) = \frac{(k+1-1)!(-1)^{k+1-1}}{x^{k+1}}$$

Proof of conjecture:

By utilizing induction hypothesis:

$$\frac{d^{k+1}}{dx^{k+1}} \ln(x) = \frac{d}{dx} \frac{(k-1)!(-1)^{k-1}}{x^k}$$

$$=\frac{d}{dx}(k-1)!(-1)^{k-1}x^{-k}$$

$$= ((k-1)!(-1)^{k-1})(\frac{d}{dx}x^{-k})$$

$$= ((k-1)!(-1)^{k-1})(-kx^{-(k+1)})$$

$$= \frac{ -k(k-1)!(-1)^{k-1} }{ x^{k+1} }$$

Not quite sure if this is correct since not getting to the desired end result ? which should be:

$$= \frac{(k+1-1)!(-1)^{k+1-1}}{x^{k+1}}$$

• – mrtaurho Oct 17 '18 at 20:59
• $(1)^{n-1}$ doesn't make sense. You should check the problem to make sure it isn't $(-1)^{n-1}$. – Acccumulation Oct 17 '18 at 21:14
• @Winther yes there is typo. It should be fixed. It didn't alter the solution i had much (or at all). – Tuki Oct 17 '18 at 21:19
• Stick the $-1$ from the front to the $(-1)^{k+1}$ term, and notice that $k(k-1)! = k! = (k+1-1)!$ – Omar Haque Oct 17 '18 at 21:25
• Possible duplicate of Conjecture for $\ln(x)$ and $\frac{d^ny}{dx^n}$ – Parcly Taxel Oct 18 '18 at 0:28

You correctly arrive at $$\frac{ -k(k-1)!\,(-1)^{k-1} }{ x^{k+1} }$$ If you plug in $$k+1$$ in the desired formula, you get $$\frac{k!\,(-1)^k}{x^{k+1}}$$ and the two formulas are actually the same: write $$-k=(-1)k$$, so $$-k(k-1)!\,(-1)^{k-1}=k(k-1)!\,(-1)(-1)^{k-1}=k!\,(-1)^k$$ as you wished.
Suppose your statement it's true for $$k\geq 1$$, then
$$\frac{d^{k+1}}{dx^{k+1}}\ln(x)=\frac{d}{dx}\left(\frac{d^{k}}{dx^{k}}\ln(x)\right)=\frac{d}{dx}\left(\frac{(k-1)!(-1)^{k-1}}{x^k}\right)=(k-1)!(-1)^{k-1}\frac{d}{dx}\left(\frac{1}{x^k}\right)=(k-1)!(-1)^{k-1}(-k)x^{-k-1}=\frac{(-k)(k-1)!(-1)^{k-1}}{x^{k+1}}=\frac{k(k-1)!\,\,(-1)(-1)^{k-1}}{x^{k+1}}=\frac{k!\,\,(-1)^{k-1+1}}{x^{k+1}}=\frac{(k+1-1)!\,\,(-1)^{k+1-1}}{x^{k+1}}.$$
• This is all in the OP, except your last $=$. – mr_e_man Oct 17 '18 at 22:27