偏微分の基礎

■問題

次の関数を偏微分せよ．

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

■答

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

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

■ヒント

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

定義域は，${\displaystyle \frac{x}{y}\ge 0}$ より，${\displaystyle \left\{\begin{array}{l}x\ge 0\\ y>0\end{array}\right.}$ および${\displaystyle \left\{\begin{array}{l}x\le 0\\ y<0\end{array}\right.}$である．

■解説

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

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

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

${\displaystyle v=\frac{x}{y}}$

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

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

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

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

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

以上より

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

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

となる．

【参考】

${\displaystyle \left\{\begin{array}{l}x\le 0\\ y<0\end{array}\right.}$の場合，以下の微分は注意が必要である．

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

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

　

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最終更新日： 2023年9月5日