微分(3)の1．(3)の解答

$3$\displaystyle{ y={ \tan }^{ - 1 }2x }$
$3$\displaystyle{\begin{array}u&=& 2x & \vspace{6}\\ \end{array}}$
$3$\displaystyle{\begin{array}y&=& { \tan }^{ - 1 }u & \vspace{6}\\ \end{array}}$ とおく。
$3$\displaystyle{\begin{array} \frac{\hspace{2} du \hspace{2}}{\hspace{2} dx \hspace{2}} &=&2& \vspace{6}\\ \end{array}}$ かつ
$3$\displaystyle{\begin{array} \frac{\hspace{2} dy \hspace{2}}{\hspace{2} du \hspace{2}} &=& \frac{\hspace{2}1\hspace{2}}{\hspace{2} 1+{u}^{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}\\ &=& 2\cdot \frac{\hspace{2}1\hspace{2}}{\hspace{2} 1+{u}^{2} \hspace{2}} & \vspace{6}\\ &=& \frac{\hspace{2}2\hspace{2}}{\hspace{2} 1+4{x}^{2} \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