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Question Number 54663 by Mikael_Marshall last updated on 08/Feb/19
y = cos2t×sen2t    y′ = ?
$${y}\:=\:{cos}\mathrm{2}{t}×{sen}\mathrm{2}{t} \\ $$$$ \\ $$$${y}'\:=\:? \\ $$
Commented by maxmathsup by imad last updated on 08/Feb/19
y(t)=cos(2t)sin(2t) ⇒y^′ (t)=−2sin^2 (2t) +2cos^2 (2t)=2(cos^2 (2t)−sin^2 (2t))  =2 cos(4t) .  another way  we have y(t)=(1/2)sin(4t) ⇒y^′ (t)=(4/2)cos(4t)=2cos(4t) .
$${y}\left({t}\right)={cos}\left(\mathrm{2}{t}\right){sin}\left(\mathrm{2}{t}\right)\:\Rightarrow{y}^{'} \left({t}\right)=−\mathrm{2}{sin}^{\mathrm{2}} \left(\mathrm{2}{t}\right)\:+\mathrm{2}{cos}^{\mathrm{2}} \left(\mathrm{2}{t}\right)=\mathrm{2}\left({cos}^{\mathrm{2}} \left(\mathrm{2}{t}\right)−{sin}^{\mathrm{2}} \left(\mathrm{2}{t}\right)\right) \\ $$$$=\mathrm{2}\:{cos}\left(\mathrm{4}{t}\right)\:. \\ $$$${another}\:{way}\:\:{we}\:{have}\:{y}\left({t}\right)=\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\mathrm{4}{t}\right)\:\Rightarrow{y}^{'} \left({t}\right)=\frac{\mathrm{4}}{\mathrm{2}}{cos}\left(\mathrm{4}{t}\right)=\mathrm{2}{cos}\left(\mathrm{4}{t}\right)\:. \\ $$
Answered by peter frank last updated on 08/Feb/19
y^′ =2cos 4t
$${y}^{'} =\mathrm{2cos}\:\mathrm{4}{t} \\ $$

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