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

Mar 6, 2018

$y ' = \frac{1 - \ln \left(3 x\right)}{3 {x}^{2}}$

#### Explanation:

$y = \frac{\ln \left(3 x\right)}{3 x}$

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

$y = \frac{1}{3} {x}^{- 1} \ln \left(3 x\right)$

$y ' = \frac{1}{3} \left(- {x}^{- 2} \ln \left(3 x\right) + {x}^{- 1} \frac{1}{x}\right)$

$y ' = \frac{1}{3} \left(\frac{- \ln \left(3 x\right)}{{x}^{2}} + \frac{1}{x} ^ 2\right)$

$y ' = \frac{1 - \ln \left(3 x\right)}{3 {x}^{2}}$

Note in the first line of taking the derivative that $\frac{d \left(\ln \left(3 x\right)\right)}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left(\ln \left(3\right)\right) + \frac{d}{\mathrm{dx}} \left(\ln \left(x\right)\right) = 0 + \frac{1}{x} = \frac{1}{x}$