How do you find the derivative of $ln(x/3)$ ?

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Answer 1 · 2017-01-26T08:01:01

$1/x$

method 1
use the chain rule

$(dy)/(dx)=(dy)/(du)xx(du)/(dx)$

$y=ln(x/3)$

$u=x/3=>(du)/(dx)=1/3$

$y=lnu=>(dy)/(du)=1/u$

$:.(dy)/(dx)=1/uxx1/3=1/(x/3)xx1/3$

$:.(dy)/(dx)=3/x xx1/3=1/x$

method 2

use rules of logs

$y=ln(x/3)=lnx-ln3$

$(dy)/(dx)=d/(dx)(lnx)-d/(dx)(ln3)$

$=1/x-0=1/x$

How do you find the derivative of ln(x/3)ln(x/3)?