微分(2)の2．(4)の解答

$3$\displaystyle{ y=\sqrt{ 2{x}^{2}- 3 } }$
$3$\begin{array}u&=& 2{x}^{2}- 3 & \vspace{6}\\ \end{array}$ とすると
$3$\displaystyle{\begin{array}y&=& \sqrt{u} & \vspace{6}\\ \end{array}}$ であるので
$3$\displaystyle{\begin{array} \frac{\hspace{2} dy \hspace{2}}{\hspace{2} du \hspace{2}} &=& \frac{\hspace{2}1\hspace{2}}{\hspace{2} 2\sqrt{u} \hspace{2}} & \vspace{6}\\ \end{array}}$ かつ $3$\displaystyle{\begin{array} \frac{\hspace{2} du \hspace{2}}{\hspace{2} dx \hspace{2}} &=& 4x & \vspace{6}\\ \end{array}}$ より
$3$\displaystyle{ \begin{array}{lll} \frac{\hspace{2} dy \hspace{2}}{\hspace{2} dx \hspace{2}} &=& \frac{\hspace{2} dy \hspace{2}}{\hspace{2} du \hspace{2}}\cdot \frac{\hspace{2} du \hspace{2}}{\hspace{2} dx \hspace{2}} & \vspace{6}\\ &=& \frac{\hspace{2}1\hspace{2}}{\hspace{2} 2\sqrt{u} \hspace{2}}\cdot 4x & \vspace{6}\\ &=& \frac{\hspace{2} 2x \hspace{2}}{\hspace{2} \sqrt{ 2{x}^{2}- 3 } \hspace{2}} & \vspace{6}\\ \end{array} }$

⇒ヒント1,ヒント2
⇒解説1
⇒微分(2)のTOPへ戻る