?@微分	$3$\displaystyle{ \log $ x+\sqrt{ {x}^{2}+A } $ }$

$3$\displaystyle{ { \{ \log \| x+\sqrt{ {x}^{2}+A } \| \} }^{\prime } }$
$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2} x+\sqrt{ {x}^{2}+A } \hspace{2}}{ $ x+\sqrt{ {x}^{2}+A } $ }^{\prime }}$
$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2} x+\sqrt{ {x}^{2}+A } \hspace{2}} \{ 1+\frac{\hspace{2} { $ {x}^{2}+A $ }^{ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} } \hspace{2}}{\hspace{2}2\hspace{2}}\cdot 2x \} }$
$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2} x+\sqrt{ {x}^{2}+A } \hspace{2}} $ 1+\frac{\hspace{2}x\hspace{2}}{\hspace{2} \sqrt{ {x}^{2}+A } \hspace{2}} $ }$
$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2} x+\sqrt{ {x}^{2}+A } \hspace{2}}\cdot \frac{\hspace{2} x+\sqrt{ {x}^{2}+A } \hspace{2}}{\hspace{2} \sqrt{ {x}^{2}+A } \hspace{2}}}$
$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{ {x}^{2}+A } \hspace{2}}}$

戻る

[ひ]
[は行]
[索引トップ]