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This question already has an answer here:

Conjecture for $\ln(x)$ and $\frac{d^ny}{dx^n}$ showed how to find the conjecture for $ln(x)$ after testing multiple values.

For this question that I have, I have to prove using induction that $$\frac{d^ny}{dx^n} (\ln(x)) = \frac{(-1)^{n-1}(n-1)!}{x^n}$$ for $n\geq1$.

I proved the base case, where $n=1$.

Then I tried to start the induction step, assuming the claim holds for some $n\geq1$. I realized that I need to show that $$\frac{d^{n+1}y}{dx^{n+1}} (\ln(x)) = \frac{(-1)^{n}(n)!}{x^{n+1}},$$ but trying to continue from here has been tricky.

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marked as duplicate by peterwhy, Cesareo, Lord Shark the Unknown, Key Flex, Parcly Taxel Oct 16 '18 at 23:08

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  • $\begingroup$ You should remove the $y$, and why is the induction step tricky? Do you know the derivative of $x^{-n}?$ $\endgroup$ – gammatester Oct 13 '18 at 19:54
  • $\begingroup$ From the way that I was taught proofs of induction, it seemed like I couldn't just find the second derivative and manipulate the right hand side of the equation so that it would equal the second derivative. $\endgroup$ – Claire Oct 13 '18 at 19:56
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Differentiate $$\frac {(-1)^{n-1}(n-1)!}{x^n}$$

Note that $$(-1)^{n-1}(n-1)!$$ is a constant and derivative of $\frac {1}{x^n}=x^{-n}$ is $-nx^{-n-1} = \frac {-n}{x^{n+1}}$

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

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

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Hint: $\frac{d^{n+1}}{dx^{n+1}} \ln x = \frac{d}{dx}\left( \frac{d^n}{dx^n} \ln x\right)$. Now use the inductive hypothesis.

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  • $\begingroup$ It should be $\frac{d^n}{dx^n}\ln x$ within the parentheses. $\endgroup$ – mrtaurho Oct 13 '18 at 20:00
  • $\begingroup$ That's what I thought @mrtaurho $\endgroup$ – Claire Oct 13 '18 at 20:03
  • $\begingroup$ So when you do this, is it technically the end of the induction step? Because all that had to be done was manipulate the equation so that you could show that $\frac{(-1)^{n-1}(n-1)!}{x^n}$=$\frac{d}{dx}\left( \frac{d^n}{dx^n} \ln x\right)$=$\frac{(-1)^{n}(n)!}{x^{n+1}}$ $\endgroup$ – Claire Oct 13 '18 at 20:12
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Since \begin{align} \frac{d}{dx} (\ln(x)) &= \frac{1}{x} \\ \frac{d^{2}}{dx^2} (\ln(x)) &= - \frac{1}{x^2} \\ \frac{d^3}{dx^3} (\ln(x)) &= \frac{(-1)^2 \, 2!}{x^3} \end{align} then is can be suggested that $$\frac{d^{n}}{dx^{n}} (\ln(x)) = \frac{(-1)^{n-1} \, (n-1)!}{x^{n}}.$$

Now, for $D = \frac{d}{dx}$, \begin{align} D^{n+1} \, \ln(x) &= D \left(D^n \, \ln(x) \right) = D \left(\frac{(-1)^{n-1} \, (n-1)!}{x^{n}} \right) \\ &= \frac{(-1)^{n} \, n!}{x^{n+1}}. \end{align} This shows that the ${(n+1)^{th}}$ term has the same form.

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