偏微分の基礎

■問題

次の関数を偏微分せよ．

$3$\displaystyle{ z={ \tan }^{ - 1 }\sqrt{ \frac{\hspace{2}x\hspace{2}}{\hspace{2}y\hspace{2}} } }$

■答

$3$\displaystyle{\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}x \hspace{2}}=\frac{\hspace{2}1\hspace{2}}{\hspace{2} 2\left({ x+y }\right)\sqrt{ \frac{\hspace{2}x\hspace{2}}{\hspace{2}y\hspace{2}} } \hspace{2}}}$

$3$\displaystyle{\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}y \hspace{2}}=- \frac{\hspace{2} \sqrt{ \frac{\hspace{2}x\hspace{2}}{\hspace{2}y\hspace{2}} } \hspace{2}}{\hspace{2} 2\left({ x+y }\right) \hspace{2}}}$

■ヒント

合成関数の微分を用いて偏微分する．
微分の際は $3$\displaystyle{{\tan }^{ - 1 }x}$ の微分の公式を用いる．

定義域は， $3$\displaystyle{\frac{\hspace{2}x\hspace{2}}{\hspace{2}y\hspace{2}}\geq 0}$ より， $3$\displaystyle{\left{{\begin{array}{lll}x\geq 0& \vspace{6}\\ y> 0& \vspace{6}\\ \end{array}}$ および $3$\displaystyle{\left{{\begin{array}{lll}x\leq 0& \vspace{6}\\ y< 0& \vspace{6}\\ \end{array}}$ である．

■解説

$3$\displaystyle{z={\tan }^{ - 1 }\sqrt{ \frac{\hspace{2}x\hspace{2}}{\hspace{2}y\hspace{2}} }}$ である．を以下のような合成関数と考える．

$3$\displaystyle{z={\tan }^{ - 1 }u}$

$3$\displaystyle{u=\sqrt{v}={v}^{ \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} }}$

$3$\displaystyle{v=\frac{\hspace{2}x\hspace{2}}{\hspace{2}y\hspace{2}}}$

$3$\displaystyle{\frac{\hspace{2} dz \hspace{2}}{\hspace{2} du \hspace{2}}}$ ， $3$\displaystyle{\frac{\hspace{2} du \hspace{2}}{\hspace{2} dv \hspace{2}}}$ ， $3$\displaystyle{\frac{\hspace{2} {\partial}v \hspace{2}}{\hspace{2} {\partial}x \hspace{2}}}$ ， $3$\displaystyle{\frac{\hspace{2} {\partial}v \hspace{2}}{\hspace{2} {\partial}y \hspace{2}}}$ を計算する．

$3$\displaystyle{\frac{\hspace{2} dz \hspace{2}}{\hspace{2} du \hspace{2}}=\frac{\hspace{2}1\hspace{2}}{\hspace{2} 1+{u}^{2} \hspace{2}}}$

$3$\displaystyle{\frac{\hspace{2} du \hspace{2}}{\hspace{2} dv \hspace{2}}=\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}{v}^{ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} }=\frac{\hspace{2}1\hspace{2}}{\hspace{2} 2\sqrt{v} \hspace{2}}}$

$3$\displaystyle{\frac{\hspace{2} {\partial}v \hspace{2}}{\hspace{2} {\partial}x \hspace{2}}=\frac{\hspace{2}1\hspace{2}}{\hspace{2}y\hspace{2}}}$

$3$\displaystyle{\frac{\hspace{2} {\partial}v \hspace{2}}{\hspace{2} {\partial}y \hspace{2}}=- \frac{\hspace{2}x\hspace{2}}{\hspace{2} {y}^{2} \hspace{2}}}$

以上より

$3$\displaystyle{\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}x \hspace{2}}=\frac{\hspace{2} dz \hspace{2}}{\hspace{2} du \hspace{2}}\frac{\hspace{2} du \hspace{2}}{\hspace{2} dv \hspace{2}}\frac{\hspace{2} {\partial}v \hspace{2}}{\hspace{2} {\partial}x \hspace{2}}}$ $3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2} 1+{u}^{2} \hspace{2}}\frac{\hspace{2}1\hspace{2}}{\hspace{2} 2\sqrt{v} \hspace{2}}\frac{\hspace{2}1\hspace{2}}{\hspace{2}y\hspace{2}}}$ $3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2} 1+\frac{\hspace{2}x\hspace{2}}{\hspace{2}y\hspace{2}} \hspace{2}}\frac{\hspace{2}1\hspace{2}}{\hspace{2} 2\sqrt{ \frac{\hspace{2}x\hspace{2}}{\hspace{2}y\hspace{2}} } \hspace{2}}\frac{\hspace{2}1\hspace{2}}{\hspace{2}y\hspace{2}}}$ $3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2} 2\left({ x+y }\right)\sqrt{ \frac{\hspace{2}x\hspace{2}}{\hspace{2}y\hspace{2}} } \hspace{2}}}$

$3$\displaystyle{\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}y \hspace{2}}=\frac{\hspace{2} dz \hspace{2}}{\hspace{2} du \hspace{2}}\frac{\hspace{2} du \hspace{2}}{\hspace{2} dv \hspace{2}}\frac{\hspace{2} {\partial}v \hspace{2}}{\hspace{2} {\partial}y \hspace{2}}}$ $3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2} 1+{u}^{2} \hspace{2}}\frac{\hspace{2}1\hspace{2}}{\hspace{2} 2\sqrt{v} \hspace{2}}\left({ - \frac{\hspace{2}x\hspace{2}}{\hspace{2} {y}^{2} \hspace{2}} }\right)}$ $3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2} 1+\frac{\hspace{2}x\hspace{2}}{\hspace{2}y\hspace{2}} \hspace{2}}\frac{\hspace{2}1\hspace{2}}{\hspace{2} 2\sqrt{ \frac{\hspace{2}x\hspace{2}}{\hspace{2}y\hspace{2}} } \hspace{2}}\left({ - \frac{\hspace{2}x\hspace{2}}{\hspace{2} {y}^{2} \hspace{2}} }\right)}$ $3$\displaystyle{=- \frac{\hspace{2} \sqrt{ \frac{\hspace{2}x\hspace{2}}{\hspace{2}y\hspace{2}} } \hspace{2}}{\hspace{2} 2\left({ x+y }\right) \hspace{2}}}$

となる．

【参考】

$3$\displaystyle{\left{{\begin{array}{lll}x\leq 0& \vspace{6}\\ y< 0& \vspace{6}\\ \end{array}}$ の場合，以下の微分は注意が必要である．

$3$\displaystyle{\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}}\sqrt{ \frac{\hspace{2}x\hspace{2}}{\hspace{2}y\hspace{2}} }=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}}\sqrt{ \frac{\hspace{2} - x \hspace{2}}{\hspace{2} - y \hspace{2}} }}$ $3$\displaystyle{=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}}\frac{\hspace{2} \sqrt{ - x } \hspace{2}}{\hspace{2} \sqrt{ - y } \hspace{2}}}$ = 1 −y 1 2 1 −x −1 $3$\displaystyle{=- \frac{\hspace{2}1\hspace{2}}{\hspace{2} 2\sqrt{ xy } \hspace{2}}}$ $3$\displaystyle{=- \frac{\hspace{2}1\hspace{2}}{\hspace{2} 2\sqrt{ - x }\sqrt{ - y } \hspace{2}}}$ $3$\displaystyle{=- \frac{\hspace{2}1\hspace{2}}{\hspace{2} 2\sqrt{ - x }\sqrt{ - y }\frac{\hspace{2} \sqrt{ - y } \hspace{2}}{\hspace{2} \sqrt{ - y } \hspace{2}} \hspace{2}}}$ $3$\displaystyle{=- \frac{\hspace{2}1\hspace{2}}{\hspace{2} 2\left({ - y }\right)\sqrt{ \frac{\hspace{2} - x \hspace{2}}{\hspace{2} - y \hspace{2}} } \hspace{2}}}$ $3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2} 2y\sqrt{ \frac{\hspace{2}x\hspace{2}}{\hspace{2}y\hspace{2}} } \hspace{2}}}$

$3$\displaystyle{\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}}\sqrt{ \frac{\hspace{2}x\hspace{2}}{\hspace{2}y\hspace{2}} }=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}}\sqrt{ \frac{\hspace{2} - x \hspace{2}}{\hspace{2} - y \hspace{2}} }}$ $3$\displaystyle{=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}}\frac{\hspace{2} \sqrt{ - x } \hspace{2}}{\hspace{2} \sqrt{ - y } \hspace{2}}}$ $3$\displaystyle{=\sqrt{ - x }\left({ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} }\right)\frac{\hspace{2}1\hspace{2}}{\hspace{2} - y\sqrt{ - y } \hspace{2}}\left({ - 1 }\right)}$ $3$\displaystyle{=- \frac{\hspace{2} \sqrt{ - x } \hspace{2}}{\hspace{2} 2y\sqrt{ - y } \hspace{2}}}$ $3$\displaystyle{=- \frac{\hspace{2}1\hspace{2}}{\hspace{2} 2y \hspace{2}}\sqrt{ \frac{\hspace{2}x\hspace{2}}{\hspace{2}y\hspace{2}} }}$ $3$\displaystyle{=- \frac{\hspace{2}1\hspace{2}}{\hspace{2} 2y \hspace{2}}\sqrt{ \frac{\hspace{2}x\hspace{2}}{\hspace{2}y\hspace{2}} }\frac{\hspace{2} \sqrt{ \frac{\hspace{2}x\hspace{2}}{\hspace{2}y\hspace{2}} } \hspace{2}}{\hspace{2} \sqrt{ \frac{\hspace{2}x\hspace{2}}{\hspace{2}y\hspace{2}} } \hspace{2}}}$ $3$\displaystyle{=- \frac{\hspace{2}1\hspace{2}}{\hspace{2} 2y \hspace{2}}\frac{\hspace{2}x\hspace{2}}{\hspace{2}y\hspace{2}}\frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{ \frac{\hspace{2}x\hspace{2}}{\hspace{2}y\hspace{2}} } \hspace{2}}}$ $3$\displaystyle{=- \frac{\hspace{2}x\hspace{2}}{\hspace{2} 2{y}^{2}\sqrt{ \frac{\hspace{2}x\hspace{2}}{\hspace{2}y\hspace{2}} } \hspace{2}}}$

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最終更新日： 2023年9月5日