Ok so I have another logarithmic derivative equation

y= ln (3x/tan x)

Normally I would do

y1= ln 3x-ln tanx

but i am thinking since there are trig functions, it has to be set up differently. Problems in the book are not helping me answer this problem.

I'm not sure what yuo mean by the term "y1" - please clarify. Do you mean deribcvitive of y?

It is true that if ln(a/b) = ln(a) - ln(b), so if y(x)= ln(3x/tanx) then y(x) = ln(3x) - ln(tanx). But now you must find the derivatives of both of these terms.