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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:

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

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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.

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Suppose your statement it's true for $k\geq 1$, then

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

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  • $\begingroup$ This is all in the OP, except your last $=$. $\endgroup$ – mr_e_man Oct 17 '18 at 22:27

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