2次の偏微分

■問題

次の関数の第2次偏導関数を求めよ．

${\displaystyle z=\sqrt{x^2-y^3}}$

■答

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

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

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

■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．

■解説

与式を変形すると，

${\displaystyle z=\sqrt{x^2-y^3}}$

平方根を累乗根の指数に変形する． 指数が有理数の場合も参照．

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

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

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

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

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

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

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

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

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

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

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

${\displaystyle =-3y{\left(x^2-y^3\right)}^{-\frac{1}{2}}-\frac{9}{4}{y}^4{\left(x^2-y^3\right)}^{-\frac{3}{2}}}$

${\displaystyle =-\frac{4\cdot 3y\left(x^2-y^3\right)+9y^4}{4\left(x^2-y^3\right)\sqrt{x^2-y^3}}}$

${\displaystyle =-\frac{12x^{2}y-12y^4+9y^4}{4\left(x^2-y^3\right)\sqrt{x^2-y^3}}}$

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

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

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

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

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

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

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

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

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

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

${\displaystyle =\frac{3x{y}^2}{2\left(x^2-y^3\right)\sqrt{x^2-y^3}}}$　･･････(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31日