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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．

${\displaystyle z={\sin }^{-1}xy}$

■答

${\displaystyle \frac{{\partial }^{2}z}{\partial x^2}=\frac{x{y}^3}{\left(1-x{}^{2}y{}^2\right)\sqrt{1-x{}^{2}y{}^2}}}$

${\displaystyle \frac{{\partial }^{2}z}{\partial y^2}=\frac{x^{3}y}{\left(1-x{}^{2}y{}^2\right)\sqrt{1-x{}^{2}y{}^2}}}$

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

■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

${\displaystyle z={\sin }^{-1}xy}$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．

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

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

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

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

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

${\displaystyle z={\sin }^{-1}xy}$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．

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

${\displaystyle =\frac{1}{\sqrt{1-{\left(xy\right)}^2}}\times \frac{\partial }{\partial x}\left(xy\right)}$

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

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

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

${\displaystyle =-\frac{1}{2}y{\left(1-x^2{y}^2\right)}^{-\frac{3}{2}}\cdot \left(-2x{y}^2\right)}$

${\displaystyle =\frac{x{y}^3}{\left(1-x^2{y}^2\right)\sqrt{1-x^2{y}^2}}}$

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

${\displaystyle =-\frac{1}{2}x{\left(1-x^2{y}^2\right)}^{-\frac{3}{2}}\cdot \left(-2x^{2}y\right)}$

${\displaystyle =\frac{x^{3}y}{\left(1-x^2{y}^2\right)\sqrt{1-x^2{y}^2}}}$

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

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

積の微分の公式を用いて

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

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

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

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

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

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