2次の偏微分

■問題

次の関数の第2次偏導関数を求めよ．

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

■答

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

■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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

${\displaystyle =-\frac{4xy}{{\left(x^2+y^2\right)}^2}}$　･･････(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： 2023年8月31日