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

Question

1689 views around the world

You can reuse this answer
Creative Commons License

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))#