微分(3)の2．(2)の解答

$3$\displaystyle{ y={ \sin }^{ - 1 }{x}^{3} }$
$3$\displaystyle{\begin{array}{lll}{y}^{\prime }&=& \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{ 1- { $ {x}^{3} $ }^{2} } \hspace{2}}{ $ {x}^{3} $ }^{\prime } & \vspace{6}\\ &=& \frac{\hspace{2} 3{x}^{2} \hspace{2}}{\hspace{2} \sqrt{ 1- {x}^{6} } \hspace{2}} & \vspace{6}\\ \end{array}}$
⇒次の問題の解答
⇒ヒント1
⇒微分(3)のTOPへ戻る