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

Question

2954 views around the world

You can reuse this answer
Creative Commons License

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#