FANDOM


\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}

Ad blocker interference detected!


Wikia is a free-to-use site that makes money from advertising. We have a modified experience for viewers using ad blockers

Wikia is not accessible if you’ve made further modifications. Remove the custom ad blocker rule(s) and the page will load as expected.