2次の偏微分

■問題

次の関数の第2次偏導関数を求めよ．

${\displaystyle z={tan}^{-1}\frac{y}{x}}$

■答

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

■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

与式を変形すると

${\displaystyle z}$${\displaystyle ={tan}^{-1}\frac{y}{x}}$

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

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

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

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

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

${\displaystyle =\frac{-y{x}^{-2}}{1+y^2{x}^{-2}}}$

${\displaystyle =\frac{-\frac{y}{x^2}}{1+\frac{y^2}{x^2}}}$

${\displaystyle =\frac{-y}{x^2+y^2}}$

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

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

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

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

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

${\displaystyle =\frac{x^{-1}}{1+y^2{x}^{-2}}}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

積の微分の公式を用いて,

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

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

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

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

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

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

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

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

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