Why is there this connection between Euler's number and $1/x$?

Question

I've seen plenty of easy to understand proofs showing that $\frac{dy}{dx}\ln(x)=\frac{1}{x}$ but what I'm looking for is a way to intuitively justify why $n=-1$ is an exception to the power rule for integration (besides the fact that it results in a $\frac{1}{0}$). Why is there this connection between $e$ and $1/x$?

Answer 1

Given $f(x)= a^x$ for some $a> 0$, the derivitive is given by the limit, as h goes to 0, of $\frac{a^{x+ h}- a^x}{h}$. That is equal to $\frac{a^xa^h- a^x}{h}= \left(\frac{a^h- 1}{h}\right)a^x$. The derivative is $\left(\lim_{h\to 0}\frac{a^h- 1}{h}\right)a^x$.

That is, the derivative of $f(x)= a^x$ is some NUMBER (that depends on the constant $a$, not the variable $x$) times $a^x$. In particular, if $a= 2$ that constant is $\lim_{h\to 0}\frac{2^h- 1}{h}$. For h= 0.001, that fraction is $\frac{2^{0.001}- 1}{0.001}= 0.6934$, much less than 1 so we can expect that the limit itself is less than 1. If x= 3, that constant is $\lim_{h\to 0}\frac{3^h-1}{h}$. When x= 0.001, that fraction is $\frac{3^{0.001}- 1}{0.001}= 1.0992$, larger than 1, implying that the limit itself is larger than 1.

That means that there exist some number a, between 2 and 3 such that the limit is 1. We call that number "e" so that the dervative of $f(x)= e^x$ is $1(e^x)$. (The world's easiest derivative!).

ln(x) is defined as the inverse function to $e^x$. That means that if $y= ln(x)$ then $x= e^y$. $\frac{dx}{dy}= e^y= x$ so $\frac{dy}{dx}= \frac{1}{x}$.

As for why the anti-derivative of $y= x^{-1}$ does not obey the "power law" you answered this yourself- if n=-1 then 1/(n+1)= 1/(-1+ 1)= 1/0 which is imposssible.