微分(3)の2．(4)の解答

$3$\displaystyle{y={x}^{3}\cos 2x}$
$3$\displaystyle{\begin{array}{lll}{y}^{\prime }&=& { $ {x}^{3} $ }^{\prime }\cos 2x+{x}^{3}{ $ \cos 2x $ }^{\prime } & \vspace{6}\\ &=& 3{x}^{2}\cos 2x+{x}^{3} $ - \sin 2x $ \cdot 2 & \vspace{6}\\ &=& 3{x}^{2}\cos 2x- 2{x}^{3}\sin 2x & \vspace{6}\\ \end{array}}$
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