How do you find the derivative of #sin^2(3x)/cos(2x)#?

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Answer 1 · 2018-04-13T17:36:00

#sin^2(3x)/cos(2x)#

#u = sin^2(3x)#
#v = cos(2x)#

#u' = 2 sin(3x) . 3 cos(3x)#
#u' = 6 sin(3x).cos(3x)#

#v' = - 2 sen(2x)#

#y' = (u'.v - u.v')/v^2#

#y' = [6 sin(3x).cos(3x).cos(2x) - sin^2(3x). {-2 sen(2x)} ]/cos^2(2x)#

#y' = [6 sin(3x).cos(3x).cos(2x) +2 sin^2(3x).sen(2x)]/cos^2(2x)#