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

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Answer 1 · 2017-12-11T15:05:45

$f'(t)=(t^2(3cos(2t)[tcos(3t)+sin(3t)]+2tsin^2(2t)))/cos^2(2t)$

$f(t)=(t^3sin(3t))/cos(2t)=g(t)/(h(t))$

$f'(t)=(h(t)g'(t)-h'(t)g(t))/(h(t))^2$

$g(t)=t^3sin(2t)$
$g'(t)=3t^3cos(3t)+3t^2sin(3t)=3t^2(tcos(3t)+sin(3t)$

$h(t)=cos(2t)$
$h'(t)=-2sin(2t)$

$f'(t)=(cos(2t)(3t^2(tcos(3t)+sin(3t))--2sin(2t)t^3sin(2t)))/(cos(2t))^2$

$=(3t^2cos(2t)(tcos(3t)+sin(3t))+2t^3sin^2(2t))/cos^2(2t)$

$=(t^2(3cos(2t)[tcos(3t)+sin(3t)]+2tsin^2(2t)))/cos^2(2t)$

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