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

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