# 2次の偏微分

## ■問題

$z=\frac{x-y}{x+y}$

## ■答

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

## ■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．

その際，商の微分の公式を用いる．

## ■解説

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

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

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

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

$=\frac{1\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\left(x+y\right)-\left(x-y\right)\text{\hspace{0.17em}}·\text{\hspace{0.17em}}1}{{\left(x+y\right)}^{2}}$

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

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

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

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

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

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

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

$=\frac{\left(-1\right)\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\left(x+y\right)-\left(x-y\right)\text{\hspace{0.17em}}·\text{\hspace{0.17em}}1}{{\left(x+y\right)}^{2}}$

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

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

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

$=\frac{\frac{\partial }{\partial x}2y\text{\hspace{0.17em}}·\text{\hspace{0.17em}}{\left(x+y\right)}^{2}-2y\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{\partial }{\partial x}{\left(x+y\right)}^{2}}{{\left\{{\left(x+y\right)}^{2}\right\}}^{2}}$

$=\frac{0\text{\hspace{0.17em}}·\text{\hspace{0.17em}}{\left(x-y\right)}^{2}-2y\text{\hspace{0.17em}}·\text{\hspace{0.17em}}2\left(x+y\right)}{{\left(x+y\right)}^{4}}$

$=\frac{0-2y\text{\hspace{0.17em}}·\text{\hspace{0.17em}}2\left(x+y\right)}{{\left(x+y\right)}^{4}}$

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

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

### ●$\frac{{\partial }^{2}z}{\partial {y}^{2}}$ 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{-2x}{{\left(x+y\right)}^{2}}\right\}$

$=\frac{\frac{\partial }{\partial y}\left(-2x\right)\text{\hspace{0.17em}}·\text{\hspace{0.17em}}{\left(x+y\right)}^{2}-\left(-2x\right)\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{\partial }{\partial y}{\left(x+y\right)}^{2}}{{\left\{{\left(x+y\right)}^{2}\right\}}^{2}}$

$=\frac{0\text{\hspace{0.17em}}·\text{\hspace{0.17em}}{\left(x+y\right)}^{2}-\left(-2x\right)\text{\hspace{0.17em}}·\text{\hspace{0.17em}}2\left(x+y\right)\text{\hspace{0.17em}}·\text{\hspace{0.17em}}1}{{\left(x+y\right)}^{4}}$

$=\frac{0-\left(-2x\right)\text{\hspace{0.17em}}·\text{\hspace{0.17em}}2\left(x+y\right)}{{\left(x+y\right)}^{4}}$

$=\frac{4x\left(x+y\right)}{{\left(x+y\right)}^{4}}$

$=\frac{4x}{{\left(x+y\right)}^{3}}$

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

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

$=\frac{\frac{\partial }{\partial y}2y\text{\hspace{0.17em}}·\text{\hspace{0.17em}}{\left(x+y\right)}^{2}-2y\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{\partial }{\partial y}{\left(x+y\right)}^{2}}{{\left\{{\left(x+y\right)}^{2}\right\}}^{2}}$

$=\frac{2{\left(x+y\right)}^{2}-2y\text{\hspace{0.17em}}·\text{\hspace{0.17em}}2\left(x+y\right)}{{\left(x+y\right)}^{4}}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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