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

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Answer 1 · 2018-03-06T04:43:29

$y' = (1-ln(3x))/(3x^2)$

$y = (ln(3x))/(3x)$

or equivalently (since I like to always use the product rule)

$y = 1/3x^(-1)ln(3x)$

$y' = 1/3(-x^(-2)ln(3x) + x^(-1)1/x)$

$y' = 1/3((-ln(3x))/(x^2) + 1/x^2)$

$y' = (1-ln(3x))/(3x^2)$

Note in the first line of taking the derivative that $(d(ln(3x)))/(dx) = (d)/(dx)(ln(3))+(d)/(dx)(ln(x)) = 0 + 1/x = 1/x$

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