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

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Answer 1 · 2018-04-13T17:36:00

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

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

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

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

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

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

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

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