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2次の偏微分

■問題

次の関数の第2次偏導関数を求めよ．

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

■答

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

■ヒント

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

${\displaystyle \frac{\partial z}{\partial x}}$ ，${\displaystyle \frac{\partial z}{\partial y}}$ 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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

${\displaystyle =\frac{0·{\left(x-y\right)}^2-2y·2\left(x+y\right)}{{\left(x+y\right)}^4}}$

${\displaystyle =\frac{0-2y·2\left(x+y\right)}{{\left(x+y\right)}^4}}$

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

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

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

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

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

${\displaystyle =\frac{0-\left(-2x\right)·2\left(x+y\right)}{{\left(x+y\right)}^4}}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

(3)，(4)より

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

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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最終更新日： 2023年8月30日

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