# 2次の偏微分

## ■問題

$z=sin\sqrt{xy}$

## ■答

$\frac{{\partial }^{2}z}{\partial {x}^{2}}=-\frac{\sqrt{y}}{4x\sqrt{x}}cos\sqrt{xy}-\frac{y}{4x}sin\sqrt{xy}$$\frac{{\partial }^{2}z}{\partial {y}^{2}}=-\frac{\sqrt{y}}{4x\sqrt{x}}cos\sqrt{xy}-\frac{y}{4x}sin\sqrt{xy}$$\frac{{\partial }^{2}z}{\partial y\partial x}=\frac{{\partial }^{2}z}{\partial x\partial y}=\frac{1}{4\sqrt{xy}}cos\sqrt{xy}-\frac{1}{4}sin\sqrt{xy}$

## ■ヒント

2次偏導関数$\frac{{\partial }^{2}z}{\partial {x}^{2}}$$\frac{{\partial }^{2}z}{\partial {y}^{2}}$$\frac{{\partial }^{2}z}{\partial y\partial x}$$\frac{{\partial }^{2}z}{\partial x\partial y}$ の4つを求める．

$\frac{\partial z}{\partial x}$$\frac{\partial z}{\partial y}$ を計算してから，それぞれを更に$x$$y$で偏微分する．

## ■解説

### ●$\frac{\partial z}{\partial x}$ の計算

$z=\mathrm{sin}\sqrt{xy}$偏導関数の定義より， $y$ を定数とみなして$x$ で微分する．

$\frac{\partial z}{\partial x}=\frac{\partial }{\partial x}\left(\mathrm{sin}\sqrt{xy}\right)$

$=cos\sqrt{xy}×\frac{\partial }{\partial x}\sqrt{xy}$

$=cos\sqrt{xy}×\frac{\partial }{\partial x}\left\{{\left(xy\right)}^{\frac{1}{2}}\right\}$

$=cos\sqrt{xy}×\frac{\partial }{\partial x}\left({x}^{\frac{1}{2}}{y}^{\frac{1}{2}}\right)$

$=cos\sqrt{xy}×\frac{1}{2}{x}^{-\frac{1}{2}}{y}^{\frac{1}{2}}$

$=cos\sqrt{xy}×\frac{\sqrt{y}}{2\sqrt{x}}$

$=\frac{\sqrt{y}}{2\sqrt{x}}cos\sqrt{xy}$　･･････(1)

### ●$\frac{\partial z}{\partial y}$ の計算

$z=\mathrm{sin}\sqrt{xy}$偏導関数の定義より， $x$ を定数とみなして$y$ で微分する．

$\frac{\partial z}{\partial y}=\frac{\partial }{\partial y}\left(\mathrm{sin}\sqrt{xy}\right)$

$=cos\sqrt{xy}×\frac{\partial }{\partial y}\sqrt{xy}$

$=cos\sqrt{xy}×\frac{\partial }{\partial y}\left\{{\left(xy\right)}^{\frac{1}{2}}\right\}$

$=cos\sqrt{xy}×\frac{\partial }{\partial y}\left({x}^{\frac{1}{2}}{y}^{\frac{1}{2}}\right)$

$=cos\sqrt{xy}×\frac{1}{2}{x}^{\frac{1}{2}}{y}^{-\frac{1}{2}}$

$=cos\sqrt{xy}×\frac{\sqrt{x}}{2\sqrt{y}}$

$=\frac{\sqrt{x}}{2\sqrt{y}}cos\sqrt{xy}$　･･････(2)

### ●$\frac{{\partial }^{2}z}{\partial {x}^{2}}$ の計算

(1)を更に， 偏導関数の定義より， $y$ を定数とみなして $x$ で微分する．

$\frac{{\partial }^{2}z}{\partial {x}^{2}}$$=\frac{\partial }{\partial x}\left(\frac{\partial z}{\partial x}\right)$

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

$=\frac{\partial }{\partial x}\left(\frac{\sqrt{y}}{2\sqrt{x}}\right)×cos\sqrt{xy}+\frac{\sqrt{y}}{2\sqrt{x}}×\frac{\partial }{\partial x}\left(cos\sqrt{xy}\right)$

$=\frac{\partial }{\partial x}\left(\frac{\sqrt{y}}{2}{x}^{-\frac{1}{2}}\right)×cos\sqrt{xy}+\frac{\sqrt{y}}{2\sqrt{x}}×\left(-sin\sqrt{xy}\right)×\frac{\partial }{\partial x}\left(\sqrt{xy}\right)$

$=-\frac{1}{2}×\frac{\sqrt{y}}{2}×{x}^{-\frac{3}{2}}×cos\sqrt{xy}-\frac{\sqrt{y}}{2\sqrt{x}}sin\sqrt{xy}×\frac{\partial }{\partial x}\left\{{\left(xy\right)}^{\frac{1}{2}}\right\}$

$=-\frac{\sqrt{y}}{4x\sqrt{x}}cos\sqrt{xy}-\frac{\sqrt{y}}{2\sqrt{x}}sin\sqrt{xy}×\frac{\partial }{\partial x}\left({x}^{\frac{1}{2}}{y}^{\frac{1}{2}}\right)$

$=-\frac{\sqrt{y}}{4x\sqrt{x}}cos\sqrt{xy}-\frac{\sqrt{y}}{2\sqrt{x}}sin\sqrt{xy}×\frac{1}{2}{x}^{-\frac{1}{2}}{y}^{\frac{1}{2}}$

$=-\frac{\sqrt{y}}{4x\sqrt{x}}cos\sqrt{xy}-\frac{\sqrt{y}}{2\sqrt{x}}sin\sqrt{xy}×\frac{\sqrt{y}}{2\sqrt{x}}$

$=-\frac{\sqrt{y}}{4x\sqrt{x}}cos\sqrt{xy}-\frac{y}{4x}sin\sqrt{xy}$

### ●$\frac{{\partial }^{2}z}{\partial {y}^{2}}$ の計算

(2)を更に， 偏導関数の定義より， $x$ を定数とみなして $y$ で微分する．

$\frac{{\partial }^{2}z}{\partial {y}^{2}}$$=\frac{\partial }{\partial y}\left(\frac{\partial z}{\partial y}\right)$

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

$=\frac{\partial }{\partial y}\left(\frac{\sqrt{x}}{2\sqrt{y}}\right)×cos\sqrt{xy}+\frac{\sqrt{x}}{2\sqrt{y}}×\frac{\partial }{\partial y}\left(cos\sqrt{xy}\right)$

$=\frac{\partial }{\partial y}\left(\frac{\sqrt{x}}{2}{y}^{-\frac{1}{2}}\right)×cos\sqrt{xy}+\frac{\sqrt{x}}{2\sqrt{y}}×\left(-sin\sqrt{xy}\right)×\frac{\partial }{\partial y}\left(\sqrt{xy}\right)$

$=-\frac{1}{2}×\frac{\sqrt{x}}{2}×{y}^{-\frac{3}{2}}×cos\sqrt{xy}-\frac{\sqrt{x}}{2\sqrt{y}}sin\sqrt{xy}×\frac{\partial }{\partial y}\left\{{\left(xy\right)}^{\frac{1}{2}}\right\}$

$=-\frac{\sqrt{x}}{4y\sqrt{y}}cos\sqrt{xy}-\frac{\sqrt{x}}{2\sqrt{y}}sin\sqrt{xy}×\frac{\partial }{\partial y}\left({x}^{\frac{1}{2}}{y}^{\frac{1}{2}}\right)$

$=-\frac{\sqrt{x}}{4y\sqrt{y}}cos\sqrt{xy}-\frac{\sqrt{x}}{2\sqrt{y}}sin\sqrt{xy}×\frac{1}{2}{x}^{-\frac{1}{2}}{y}^{\frac{1}{2}}$

$=-\frac{\sqrt{x}}{4y\sqrt{y}}cos\sqrt{xy}-\frac{\sqrt{x}}{2\sqrt{y}}sin\sqrt{xy}×\frac{\sqrt{x}}{2\sqrt{y}}$

$=-\frac{\sqrt{x}}{4y\sqrt{y}}cos\sqrt{xy}-\frac{x}{4y}sin\sqrt{xy}$

### ●$\frac{{\partial }^{2}z}{\partial y\partial x}$ の計算

(1)を更に，偏導関数の定義より， $x$ を定数とみなして $y$ で微分する．

$\frac{{\partial }^{2}z}{\partial y\partial x}$$=\frac{\partial }{\partial y}\left(\frac{\partial z}{\partial x}\right)$

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

$=\frac{\partial }{\partial y}\left(\frac{\sqrt{y}}{2\sqrt{x}}\right)×cos\sqrt{xy}+\frac{\sqrt{y}}{2\sqrt{x}}×\frac{\partial }{\partial y}\left(cos\sqrt{xy}\right)$

$=\frac{\partial }{\partial y}\left(\frac{1}{2\sqrt{x}}{y}^{\frac{1}{2}}\right)×cos\sqrt{xy}+\frac{\sqrt{y}}{2\sqrt{x}}×\left(-sin\sqrt{xy}\right)×\frac{\partial }{\partial y}\left(\sqrt{xy}\right)$

$=\frac{1}{2}×\frac{1}{2\sqrt{x}}×{y}^{-\frac{1}{2}}×cos\sqrt{xy}-\frac{\sqrt{y}}{2\sqrt{x}}sin\sqrt{xy}×\frac{\partial }{\partial y}\left\{{\left(xy\right)}^{\frac{1}{2}}\right\}$

$=\frac{1}{4\sqrt{xy}}cos\sqrt{xy}-\frac{\sqrt{y}}{2\sqrt{x}}sin\sqrt{xy}×\frac{\partial }{\partial y}\left({x}^{\frac{1}{2}}{y}^{\frac{1}{2}}\right)$

$=\frac{1}{4\sqrt{xy}}cos\sqrt{xy}-\frac{\sqrt{y}}{2\sqrt{x}}sin\sqrt{xy}×\frac{1}{2}{x}^{\frac{1}{2}}{y}^{-\frac{1}{2}}$

$=\frac{1}{4\sqrt{xy}}cos\sqrt{xy}-\frac{\sqrt{y}}{2\sqrt{x}}sin\sqrt{xy}×\frac{\sqrt{x}}{2\sqrt{y}}$

$=\frac{1}{4\sqrt{xy}}cos\sqrt{xy}-\frac{\sqrt{xy}}{4\sqrt{xy}}sin\sqrt{xy}$

$=\frac{1}{4\sqrt{xy}}cos\sqrt{xy}-\frac{1}{4}sin\sqrt{xy}$　･･････(3)

### ●$\frac{{\partial }^{2}z}{\partial x\partial y}$ の計算

(2)を更に，偏導関数の定義より， $y$ を定数とみなして $x$ で微分する．

$\frac{{\partial }^{2}z}{\partial x\partial y}$$=\frac{\partial }{\partial x}\left(\frac{\partial z}{\partial y}\right)$

$=\frac{\partial z}{\partial x}\left(\frac{\sqrt{x}}{2\sqrt{y}}\mathrm{cos}\sqrt{xy}\right)$

$=\frac{\partial z}{\partial x}\left(\frac{\sqrt{x}}{2\sqrt{y}}\right)×\mathrm{cos}\sqrt{xy}+\frac{\sqrt{x}}{2\sqrt{y}}×\left(-\mathrm{sin}\sqrt{xy}\right)×\frac{\partial }{\partial x}\left(\sqrt{xy}\right)$

$=\frac{1}{2}×\frac{1}{2\sqrt{y}}×\frac{1}{\sqrt{x}}×\mathrm{cos}\sqrt{xy}+\frac{\sqrt{x}}{2\sqrt{y}}×\left(-\mathrm{sin}\sqrt{xy}\right)×\frac{1}{2}\cdot \frac{\sqrt{y}}{\sqrt{x}}$

$=\frac{1}{4\sqrt{xy}}\mathrm{cos}\sqrt{xy}-\frac{1}{4}\mathrm{sin}\sqrt{xy}$　･･････(4)

(3)，(4)より

$\frac{{\partial }^{2}z}{\partial y\partial x}=\frac{{\partial }^{2}z}{\partial x\partial y}$

となり，2次偏導関数は偏微分する順序には無関係であることが確かめられた．

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