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

Question

2130 views around the world

You can reuse this answer
Creative Commons License

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)#