How do you differentiate $(ln(2x) )/ (cos2x)$ using the quotient rule?

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Answer 1 · 2016-01-11T00:27:14

$f'(x)=(cos(2x)+2xln(2x)*sin(2x))/(xcos^2(2x))$

$f(x)=(h(x))/g(x)$

with:

$h(x)=ln(2x)=>h'(x)=1/(cancel(2)x)*cancel(2)=1/x$

$g(x)=cos(2x)=>g'(x)=-sin(2x)*2=-2sin(2x)$

$f'(x)=(h'(x)*g(x)-h(x)*g'(x))/[g(x)]^2$

then:

$f'(x)=(1/x*cos(2x)-ln(2x)*(-2sin(2x)))/[cos(2x)]^2=$

$=(1/x*cos(2x)+2*ln(2x)*sin(2x))/cos^2(2x)=$

$(cos(2x)+2xln(2x)*sin(2x))/(xcos^2(2x))$

How do you differentiate (ln(2x) )/ (cos2x)(ln(2x) )/ (cos2x) using the quotient rule?