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

■答え

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

■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 =\sqrt{2x-3y}}$

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．

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

(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．

　

ホーム>>カテゴリー分類>>微分>>偏微分>>問題演習>>2次の偏微分

学生スタッフ作成

最終更新日： 2023年8月30日