偏微分を含む証明

■問題

次のことを証明せよ．

$3$\displaystyle{z=\frac{\hspace{2}1\hspace{2}}{\hspace{2}x\hspace{2}}f $ \frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}} $ \text{\hspace{1} }}$ ならば $3$\displaystyle{x\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}x \hspace{2}}+y\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}y \hspace{2}}+z=0\text{\hspace{1} }}$ である．

■ヒント

$3$\displaystyle{ z\text{\hspace{1} } }$ を $3$\displaystyle{ x\text{\hspace{1} },\text{\hspace{1} }y\text{\hspace{1} } }$ でそれぞれ偏微分し，２式を連立させる．このとき，積の微分と合成関数の微分を用いる．

■解説

$3$\displaystyle{u=\frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}}\text{\hspace{1} }}$ とおくと

$3$\displaystyle{z=\frac{\hspace{2}1\hspace{2}}{\hspace{2}x\hspace{2}}f $u$ }$ ， $3$\displaystyle{\frac{\hspace{2} {\partial}u \hspace{2}}{\hspace{2} {\partial}x \hspace{2}}=- \frac{\hspace{2}y\hspace{2}}{\hspace{2} {x}^{2} \hspace{2}}}$ ， $3$\displaystyle{\frac{\hspace{2} {\partial}u \hspace{2}}{\hspace{2} {\partial}y \hspace{2}}=\frac{\hspace{2}1\hspace{2}}{\hspace{2}x\hspace{2}}}$

偏導関数の定義より， $3$\displaystyle{y\text{\hspace{1}}}$ を定数とみなして $3$\displaystyle{x\text{\hspace{1}}}$ で微分する（ここを参照）．

$3$\displaystyle{\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}x \hspace{2}}\text{\hspace{1}}=\left{{ \frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}}\left({ \frac{\hspace{2}1\hspace{2}}{\hspace{2}x\hspace{2}} }\right) }\right}\text{\hspace{1}}\cdot \text{\hspace{1}}f\left({u}\right)+\frac{\hspace{2}1\hspace{2}}{\hspace{2}x\hspace{2}}\text{\hspace{1}}\cdot \text{\hspace{1}}\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}}f\left({u}\right)}$

$3$\displaystyle{=\left{{ \frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}}\left({ \frac{\hspace{2}1\hspace{2}}{\hspace{2}x\hspace{2}} }\right) }\right}\text{\hspace{1}}\cdot \text{\hspace{1}}f\left({u}\right)+\frac{\hspace{2}1\hspace{2}}{\hspace{2}x\hspace{2}}\text{\hspace{1}}\cdot \text{\hspace{1}}{f}^{\prime }\left({u}\right)\text{\hspace{1}}\cdot \text{\hspace{1}}\frac{\hspace{2} {\partial}u \hspace{2}}{\hspace{2} {\partial}x \hspace{2}}}$

$3$\displaystyle{=\left({ - \frac{\hspace{2}1\hspace{2}}{\hspace{2} {x}^{2} \hspace{2}} }\right)\cdot f\left({ \frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}} }\right)+\frac{\hspace{2}1\hspace{2}}{\hspace{2}x\hspace{2}}\cdot {f}^{\prime }\left({ \frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}} }\right)\cdot \left({ - \frac{\hspace{2}y\hspace{2}}{\hspace{2} {x}^{2} \hspace{2}} }\right)}$

$3$\displaystyle{ =- \frac{\hspace{2}1\hspace{2}}{\hspace{2} {x}^{2} \hspace{2}}\cdot f $ \frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}} $ - \frac{\hspace{2}y\hspace{2}}{\hspace{2} {x}^{3} \hspace{2}}\cdot {f}^{\prime } $ \frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}} $ }$

より

$3$\displaystyle{- \frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}x \hspace{2}}- \frac{\hspace{2}1\hspace{2}}{\hspace{2} {x}^{2} \hspace{2}}f $ \frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}} $ \text{\hspace{1} }=\frac{\hspace{2}y\hspace{2}}{\hspace{2} {x}^{3} \hspace{2}}{f}^{\prime } $ \frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}} $ }$

$3$\displaystyle{- \frac{\hspace{2} {x}^{3} \hspace{2}}{\hspace{2}y\hspace{2}}\text{\hspace{1} }\cdot \text{\hspace{1} }\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}x \hspace{2}}- \frac{\hspace{2}x\hspace{2}}{\hspace{2}y\hspace{2}}f $ \frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}} $ ={f}^{\prime } $ \frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}} $ }$

偏導関数の定義より， $3$\displaystyle{x\text{\hspace{1}}}$ を定数とみなして $3$\displaystyle{y\text{\hspace{1}}}$ で微分する（ここを参照）．

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

$3$\displaystyle{ \frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}y \hspace{2}}\text{\hspace{1}}=\left{{ \frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}}\left({ \frac{\hspace{2}1\hspace{2}}{\hspace{2}x\hspace{2}} }\right) }\right}\text{\hspace{1}}\cdot \text{\hspace{1}}f\left({u}\right)+\frac{\hspace{2}1\hspace{2}}{\hspace{2}x\hspace{2}}\text{\hspace{1}}\cdot \text{\hspace{1}}{f}^{\prime }\left({u}\right)\text{\hspace{1}}\cdot \text{\hspace{1}}\frac{\hspace{2} {\partial}u \hspace{2}}{\hspace{2} {\partial}y \hspace{2}} }$

$3$\displaystyle{ =0\cdot f $u$ +\frac{\hspace{2}1\hspace{2}}{\hspace{2}x\hspace{2}}\cdot {f}^{\prime } $ \frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}} $ \cdot \frac{\hspace{2}1\hspace{2}}{\hspace{2}x\hspace{2}} }$

$3$\displaystyle{ =\frac{\hspace{2}1\hspace{2}}{\hspace{2} {x}^{2} \hspace{2}}\cdot {f}^{\prime } $ \frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}} $ }$

より

$3$\displaystyle{{x}^{2}\text{\hspace{1} }\cdot \text{\hspace{1} }\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}y \hspace{2}}={f}^{\prime } $ \frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}} $ }$

以上から

$3$\displaystyle{- \frac{\hspace{2} {x}^{3} \hspace{2}}{\hspace{2}y\hspace{2}}\text{\hspace{1} }\cdot \text{\hspace{1} }\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}x \hspace{2}}- \frac{\hspace{2}x\hspace{2}}{\hspace{2}y\hspace{2}}f $ \frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}} $ \text{\hspace{1} }={x}^{2}\text{\hspace{1} }\cdot \text{\hspace{1} }\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}y \hspace{2}}}$

$3$\displaystyle{\frac{\hspace{2} {x}^{3} \hspace{2}}{\hspace{2}y\hspace{2}}\text{\hspace{1} }\cdot \text{\hspace{1} }\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}x \hspace{2}}+{x}^{2}\text{\hspace{1} }\cdot \text{\hspace{1} }\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}y \hspace{2}}+\frac{\hspace{2}x\hspace{2}}{\hspace{2}y\hspace{2}}f $ \frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}} $ \text{\hspace{1} }=0}$

両辺に $3$\displaystyle{\frac{\hspace{2}y\hspace{2}}{\hspace{2} {x}^{2} \hspace{2}}\text{\hspace{1} }}$ をかける．

$3$\displaystyle{x\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}x \hspace{2}}+y\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}y \hspace{2}}+\frac{\hspace{2}1\hspace{2}}{\hspace{2}x\hspace{2}}f $ \frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}} $ \text{\hspace{1} }=0}$

$3$\displaystyle{z=\frac{\hspace{2}1\hspace{2}}{\hspace{2}x\hspace{2}}f $ \frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}} $ \text{\hspace{1} }}$ を代入する．

$3$\displaystyle{x\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}x \hspace{2}}+y\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}y \hspace{2}}+z\text{\hspace{1} }=0}$

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最終更新日： 2023年8月25日