Đáp án:
$\begin{array}{l}
n)y = {\sin ^3}2x.{\cos ^3}2x\\
y' = \left( {{{\sin }^3}2x} \right)'.{\cos ^3}2x + {\sin ^3}2x.\left( {{{\cos }^3}2x} \right)'\\
= 2.3.\left( {\sin 2x} \right)'.{\sin ^2}2x.{\cos ^3}2x\\
+ {\sin ^3}2x.2.3.\left( {\cos 2x} \right)'.{\cos ^2}2x\\
= 6.si{n^2}2x.{\cos ^4}2x - 6.{\sin ^4}2x.{\cos ^2}2x\\
= 6{\sin ^2}2x.{\cos ^2}2x.\left( {{{\cos }^2}2x - {{\sin }^2}2x} \right)\\
= \frac{3}{2}{\sin ^2}4x.\cos 4x\\
o)y = \sin \left( {\cos 3x} \right)\\
\Rightarrow y' = \left( {\cos 3x} \right)'.\cos \left( {\cos 3x} \right)\\
= - 3\sin 3x.\cos \left( {\cos 3x} \right)
\end{array}$
