How do you differentiate #f(x)=tan(3x)#?

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Answer 1 · 2016-11-04T17:04:43

#3sec^2 3x#

Using chain rule, first differentiate tan3x w.r.t 3x and then differentiate 3x w.rt x

Accordingly, f'(x) = #sec^2 3x *3# = #3sec^2 3x# Answer

Answer 2 · 2016-11-04T17:09:32

The answer is #=3sec^2(3x)#

First you have to know that #(tanx)'=(sinx/cosx)'#

# = (cosx*cosx--sinx*sinx) /(cos^2x)#

#=1/cos^2x= sec^2x# since #(cos^2x+sin^2x=1)#
#(tanx)'=sec^2x#

if #f(x)=tan(3x)#
then #f'(x)=sec^2(3x)*3# by the chain rule