# 2次の偏微分

## ■問題

$z=\mathrm{cos}{x}^{2}{y}^{2}$

## ■答

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

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

$\frac{{\partial }^{2}z}{\partial y\partial x}=\frac{{\partial }^{2}z}{\partial x\partial y}$ $=-4xy\mathrm{sin}{x}^{2}{y}^{2}-4{x}^{3}{y}^{3}\mathrm{cos}{x}^{2}{y}^{2}$

## ■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

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

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

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

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

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

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

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

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

$=-2{y}^{2}\mathrm{sin}{x}^{2}y{}^{2}-2x{y}^{2}\cdot 2x{y}^{2}\mathrm{cos}{x}^{2}y{}^{2}$

$=-2{y}^{2}\mathrm{sin}{x}^{2}y{}^{2}-4{x}^{2}{y}^{4}\mathrm{cos}{x}^{2}y{}^{2}$

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

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

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

$=-2{x}^{2}\mathrm{sin}{x}^{2}y{}^{2}-2{x}^{2}y\cdot 2{x}^{2}y\mathrm{cos}{x}^{2}y{}^{2}$

$=-2{x}^{2}\mathrm{sin}{x}^{2}y{}^{2}-4{x}^{4}{y}^{2}\mathrm{cos}{x}^{2}y{}^{2}$

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

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

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

$=-4xy\mathrm{sin}{x}^{2}{y}^{2}-2x{y}^{2}\cdot 2{x}^{2}y\mathrm{cos}{x}^{2}{y}^{2}$

$=-4xy\mathrm{sin}{x}^{2}{y}^{2}-4{x}^{3}{y}^{3}\mathrm{cos}{x}^{2}{y}^{2}$　･･････(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．

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

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

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

$=-4xy\mathrm{sin}{x}^{2}{y}^{2}-2{x}^{2}y\mathrm{cos}{x}^{2}{y}^{2}×2x{y}^{2}$

$=-4xy\mathrm{sin}{x}^{2}{y}^{2}-4{x}^{3}{y}^{3}\mathrm{cos}{x}^{2}{y}^{2}$　･･････(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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