## FANDOM

1,168 Pages

$\frac{d}{dx}\big(\ln(|x|)\big)=\frac{1}{x}$ when $x\ne0$ .

## Proof

$f(x)=\ln(|x|)$
$f'(x)=\lim_{h\to0}\frac{f(x+h) - f(x)}{\Delta x}$
$f'(x)=\lim_{h\to0}\frac{\ln(|x+h|)-\ln(|x|)}{h}$
$f'(x)=\lim_{h\to0}\frac{\ln\left(\frac{|x+h|}{|x|}\right)}{h}$
$f'(x)=\lim_{h\to0}\frac{\ln\left(\left|\frac{x+h}{x}\right|\right)}{h}$
$f'(x)=\lim_{h\to0}\frac{1}{h}\ln\left(\left|\frac{x+h}{x}\right|\right)$
$f'(x)=\lim_{h\to0}\ln\left(\left|\frac{x+h}{x}\right|^\frac{1}{h}\right)$
$f'(x)=\lim_{h\to0}\ln\left(\left|1+\frac{h}{x}\right|^\frac{1}{h}\right)$

Now, by making the substitution

$n=\frac{h}{x}$
$\lim_{h\to0}n=\lim_{h\to0}\frac{h}{x}=0$
$f'(x)=\lim_{n\to0}\ln\left((1+n)^\frac{1}{nx}\right)$
$f'(x)=\lim_{n\to0}\frac{1}{x}\ln\left((1+n)^\frac{1}{n}\right)$

One definition of Euler's number is

$\lim_{n\to0}(1+n)^\frac{1}{n}=e$

so the expression simplifies to

$f'(x)=\ln(e)\frac{1}{x}=\frac{1}{x}$
Community content is available under CC-BY-SA unless otherwise noted.