# Conjecture for $\ln(x)$ and $\frac{d^ny}{dx^n}$

I'm not really sure where to start, I found the first, second, third and fourth derivatives of $$\ln(x)$$ to be $$\frac{1}{x}$$,-$$\frac{1}{x^2}$$, $$\frac{2}{x^3}$$, and -$$\frac{6}{x^4}$$, respectively. Letting $$n$$ be a natural number, I have to formulate a conjecture for a formula for $$\frac{d^ny}{dx^n}$$. Afterward, I have to use mathematical induction to prove the conjecture.

• How about computing a couple more? – Lord Shark the Unknown Oct 13 '18 at 18:55

Hint:

$$\Bigl(\frac1{x^n}\Bigr)'=-\frac n{x^{n+1}}.$$

• So this is true for $n\geq1$, which would be included in the conjecture, and I understand where you found this from because it is true. Am I supposed to try and prove that -$\frac{n+1}{x^{(n+1)+1}}$ is true to to prove the conjecture? – Claire Oct 13 '18 at 19:06
• Maybe, depending on what you're supposed to know. Personally, I consider that every well-bred young people should know this formula by heart. – Bernard Oct 13 '18 at 19:12
• I'm just supposed to be able to figure out the conjecture and prove it via induction – Claire Oct 13 '18 at 19:12
• So I would do it in two step: first establish this formula (which is very easy) and next use it for the $n$-th derivative of log. – Bernard Oct 13 '18 at 19:16
• You'll indeed obtain factorials, with a sign. Try first to formulate a conjecture, and test it for a few more orders of derivation, to see whether it seems valid, then prove it by induction. – Bernard Oct 13 '18 at 19:23

You've probably at least guessed that a sequence $$a_n$$ of positive integers exists for which the $$n$$th derivative of $$\ln x$$ is $$(-1)^{n-1}a_n/x^n$$. What's $$a_1$$? What's $$a_{n+1}$$ in terms of $$a_n$$? Can you finish the solution using factorials?

• $a_1$ would be $(n)!$ or (1)!, and then $a_{n+1}$ would be -(1)!, but how are you able to get to (6)! by the fourth derivative? – Claire Oct 13 '18 at 19:24
• @Claire I think you meant $a_1=1=0!,\,a_{n+1}/a_n=n$. – J.G. Oct 13 '18 at 19:25
• ohhhhh i see now! – Claire Oct 13 '18 at 19:27
• okay so the conjecture could be $\frac{(-1)^{n-1}(n-1)!}{x^n}$ for n≥1? – Claire Oct 13 '18 at 19:31
• @Claire So now prove it by induction. – J.G. Oct 13 '18 at 20:55