How do you differentiate #ln(3x)#?

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Answer 1 · 2016-07-10T08:16:39

#1/x#

#d/(du) ln u = 1/u#

#u = 3x, (du)/dx = 3#

#d/(dx) ln 3x = 1/(3x) * 3 = 1/x#

Answer 2 · 2016-07-10T08:27:12

Let #y=ln(3x)#.

Since #d/(dx)[ln(u)] = (u')/(u)#, let #u = 3x#, so

#y'=3/(3x)=1/x#

In fact, we can generalize this formula even more if we notice that for any number #a#, and #y=ln(ax)#, then #y'=a/(ax)=1/x#