# 2次の偏微分

## ■問題

$z={\mathrm{sin}}^{-1}xy$

## ■答

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

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

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

## ■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

$z={\mathrm{sin}}^{-1}xy$偏導関数の定義より， $y$ を定数とみなして$x$ で微分する．

$\frac{\partial z}{\partial x}=\frac{\partial }{\partial x}\left({\mathrm{sin}}^{-1}xy\right)$

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

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

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

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

$z={\mathrm{sin}}^{-1}xy$偏導関数の定義より， $x$ を定数とみなして$y$ で微分する．

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

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

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

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

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

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

$=-\frac{1}{2}y{\left(1-{x}^{2}{y}^{2}\right)}^{-\frac{3}{2}}\cdot \left(-2x{y}^{2}\right)$

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

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

$=x\cdot \left(-\frac{1}{2}\right){\left(1-{x}^{2}{y}^{2}\right)}^{-\frac{3}{2}}×\frac{\partial }{\partial y}\left(1-{x}^{2}{y}^{2}\right)$

$=-\frac{1}{2}x{\left(1-{x}^{2}{y}^{2}\right)}^{-\frac{3}{2}}\cdot \left(-2{x}^{2}y\right)$

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

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

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

$={\left(1-{x}^{2}{y}^{2}\right)}^{-\frac{1}{2}}+y\cdot \left(-\frac{1}{2}\right){\left(1-{x}^{2}{y}^{2}\right)}^{-\frac{3}{2}}\cdot \frac{\partial }{\partial y}\left(1-{x}^{2}{y}^{2}\right)$

$={\left(1-{x}^{2}{y}^{2}\right)}^{-\frac{1}{2}}+-\frac{1}{2}y{\left(1-{x}^{2}{y}^{2}\right)}^{-\frac{3}{2}}×\left(-2{x}^{2}y\right)$

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

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

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

$={\left(1-{x}^{2}{y}^{2}\right)}^{-\frac{1}{2}}+x\cdot \left(-\frac{1}{2}\right)\cdot {\left(1-{x}^{2}{y}^{2}\right)}^{-\frac{3}{2}}×\frac{\partial }{\partial x}\left(1-{x}^{2}{y}^{2}\right)$

$={\left(1-{x}^{2}{y}^{2}\right)}^{-\frac{1}{2}}+-\frac{1}{2}x{\left(1-{x}^{2}{y}^{2}\right)}^{-\frac{3}{2}}×\left(-2x{y}^{2}\right)$

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

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