微分(3)の1．(4)の解答

$3$\displaystyle{y=\sin \frac{\hspace{2}1\hspace{2}}{\hspace{2}x\hspace{2}}}$
$3$\displaystyle{\begin{array}u&=& \frac{\hspace{2}1\hspace{2}}{\hspace{2}x\hspace{2}} & \vspace{6}\\ \end{array}}$
$3$\displaystyle{\begin{array}y&=& \sin u & \vspace{6}\\ \end{array}}$ とおく。
$3$\displaystyle{\begin{array} \frac{\hspace{2} dy \hspace{2}}{\hspace{2} du \hspace{2}} &=& \cos u & \vspace{6}\\ \end{array}}$ かつ
$3$\displaystyle{\begin{array} \frac{\hspace{2} du \hspace{2}}{\hspace{2} dx \hspace{2}} &=& - \frac{\hspace{2}1\hspace{2}}{\hspace{2} {x}^{2} \hspace{2}} & \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}\\ &=& $ \cos \frac{\hspace{2}1\hspace{2}}{\hspace{2}x\hspace{2}} $ \cdot $ - \frac{\hspace{2}1\hspace{2}}{\hspace{2} {x}^{2} \hspace{2}} $ & \vspace{6}\\ &=& - \frac{\hspace{2}1\hspace{2}}{\hspace{2} {x}^{2} \hspace{2}}\cos \frac{\hspace{2}1\hspace{2}}{\hspace{2}x\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