# 2次の偏微分

## ■問題

$z={\mathrm{cos}}^{-1}2xy$

## ■答

$\frac{{\partial }^{2}z}{\partial {x}^{2}}=-\frac{8x{y}^{3}}{\left(1-4{x}^{2}{y}^{2}\right)\sqrt{1-4{x}^{2}{y}^{2}}}$

$\frac{{\partial }^{2}z}{\partial {y}^{2}}=-\frac{8{x}^{3}y}{\left(1-4{x}^{2}{y}^{2}\right)\sqrt{1-4{x}^{2}{y}^{2}}}$

$\frac{{\partial }^{2}z}{\partial y\partial x}=\frac{{\partial }^{2}z}{\partial x\partial y}$ $=-\frac{2}{\left(1-4{x}^{2}{y}^{2}\right)\sqrt{1-4{x}^{2}{y}^{2}}}$

## ■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

$z={\mathrm{cos}}^{-1}2xy$偏導関数の定義より， $y$ を定数とみなして$x$ で微分する．

$\frac{\partial z}{\partial x}=\frac{\partial }{\partial x}\left({\mathrm{cos}}^{-1}2xy\right)$

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

$=-\frac{2y}{\sqrt{1-4{x}^{2}{y}^{2}}}$

$=-2y{\left(1-4{x}^{2}{y}^{2}\right)}^{-\frac{1}{2}}$　･･････(1)

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

$z={\mathrm{cos}}^{-1}2xy$偏導関数の定義より， $x$ を定数とみなして$y$ で微分する．

$\frac{\partial z}{\partial y}=\frac{\partial }{\partial y}\left({\mathrm{cos}}^{-1}2xy\right)$

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

$=-\frac{2x}{\sqrt{1-4{x}^{2}{y}^{2}}}$

$=-2x{\left(1-4{x}^{2}{y}^{2}\right)}^{-\frac{1}{2}}$　･･････(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)$

$=\frac{\partial }{\partial x}\left\{-2y{\left(1-4{x}^{2}{y}^{2}\right)}^{-\frac{1}{2}}\right\}$

$=-2y\cdot \left(-\frac{1}{2}\right){\left(1-4{x}^{2}{y}^{2}\right)}^{-\frac{3}{2}}×\frac{\partial }{\partial x}\left(1-4{x}^{2}{y}^{2}\right)$

$=y{\left(1-4{x}^{2}{y}^{2}\right)}^{-\frac{3}{2}}\cdot \left(-8x{y}^{2}\right)$

$=-\frac{8x{y}^{3}}{\left(1-4{x}^{2}{y}^{2}\right)\sqrt{1-4{x}^{2}{y}^{2}}}$

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

$=-2x\cdot \left(-\frac{1}{2}\right){\left(1-4{x}^{2}{y}^{2}\right)}^{-\frac{3}{2}}×\frac{\partial }{\partial y}\left(1-4{x}^{2}{y}^{2}\right)$

$=x{\left(1-4{x}^{2}{y}^{2}\right)}^{-\frac{3}{2}}\cdot \left(-8{x}^{2}y\right)$

$=-\frac{8{x}^{3}y}{\left(1-4{x}^{2}{y}^{2}\right)\sqrt{1-4{x}^{2}{y}^{2}}}$

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

$=\frac{\partial }{\partial y}\left\{-2y{\left(1-4{x}^{2}{y}^{2}\right)}^{-\frac{1}{2}}\right\}$

$=-2{\left(1-4{x}^{2}{y}^{2}\right)}^{-\frac{1}{2}}+\left(-2y\right)\cdot \left(-\frac{1}{2}\right){\left(1-4{x}^{2}{y}^{2}\right)}^{-\frac{3}{2}}×\frac{\partial }{\partial y}\left(1-4{x}^{2}{y}^{2}\right)$

$=-2{\left(1-4{x}^{2}{y}^{2}\right)}^{-\frac{1}{2}}+y{\left(1-4{x}^{2}{y}^{2}\right)}^{-\frac{3}{2}}\cdot \left(-8{x}^{2}y\right)$

$=\frac{-2+8{x}^{2}{y}^{2}-8{x}^{2}{y}^{2}}{\left(1-4{x}^{2}{y}^{2}\right)\sqrt{1-4{x}^{2}{y}^{2}}}$

$=-\frac{2}{\left(1-4{x}^{2}{y}^{2}\right)\sqrt{1-4{x}^{2}{y}^{2}}}$　･･････(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 x\partial y}$$=\frac{\partial }{\partial x}\left(\frac{\partial z}{\partial y}\right)$

$=\frac{\partial }{\partial x}\left\{-2x{\left(1-4{x}^{2}{y}^{2}\right)}^{-\frac{1}{2}}\right\}$

$=-2{\left(1-4{x}^{2}{y}^{2}\right)}^{-\frac{1}{2}}+\left(-2x\right)\cdot \left(-\frac{1}{2}\right){\left(1-4{x}^{2}{y}^{2}\right)}^{-\frac{3}{2}}×\frac{\partial }{\partial x}\left(1-4{x}^{2}{y}^{2}\right)$

$=-2{\left(1-4{x}^{2}{y}^{2}\right)}^{-\frac{1}{2}}+x{\left(1-4{x}^{2}{y}^{2}\right)}^{-\frac{3}{2}}\cdot \left(-8x{y}^{2}\right)$

$=\frac{-2\left(1-4{x}^{2}{y}^{2}\right)-8x{y}^{2}}{\left(1-4{x}^{2}{y}^{2}\right)\sqrt{1-4{x}^{2}{y}^{2}}}$

$=-\frac{2}{\left(1-4{x}^{2}{y}^{2}\right)\sqrt{1-4{x}^{2}{y}^{2}}}$　･･････(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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