2次の偏微分

■問題

次の関数の第2次偏導関数を求めよ．

${\displaystyle z=\log \left(x-y\right)}$

■答

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

■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．

■解説

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

${\displaystyle z=\log \left(x-y\right)}$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．

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

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

${\displaystyle =\frac{1}{x-y}·1}$

${\displaystyle =\frac{1}{x-y}}$

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

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

${\displaystyle z=\log \left(x-y\right)}$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．

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

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

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

${\displaystyle =-\frac{1}{x-y}}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

${\displaystyle =\frac{1}{{\left(x-y\right)}^2}}$　･･････(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： 2026年9月29日