2次の偏微分

■問題

次の関数の第2次偏導関数を求めよ．

${\displaystyle z={\cos }^{-1}2xy}$

■答

${\displaystyle \frac{{\partial }^{2}z}{\partial x^2}=-\frac{8x{y}^3}{\left(1-4x^2{y}^2\right)\sqrt{1-4x^2{y}^2}}}$

${\displaystyle \frac{{\partial }^{2}z}{\partial y^2}=-\frac{8x^{3}y}{\left(1-4x^2{y}^2\right)\sqrt{1-4x^2{y}^2}}}$

${\displaystyle \frac{{\partial }^{2}z}{\partial y\partial x}=\frac{{\partial }^{2}z}{\partial x\partial y}}$ ${\displaystyle =-\frac{2}{\left(1-4x^2{y}^2\right)\sqrt{1-4x^2{y}^2}}}$

■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

${\displaystyle z={\cos }^{-1}2xy}$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．

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

${\displaystyle =-\frac{1}{\sqrt{1-4x^2{y}^2}}\times \frac{\partial }{\partial x}\left(2xy\right)}$

${\displaystyle =-\frac{2y}{\sqrt{1-4x^2{y}^2}}}$

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

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

${\displaystyle z={\cos }^{-1}2xy}$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．

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

${\displaystyle =-\frac{1}{\sqrt{1-4x^2{y}^2}}\times \frac{\partial }{\partial y}\left(2xy\right)}$

${\displaystyle =-\frac{2x}{\sqrt{1-4x^2{y}^2}}}$

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

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

${\displaystyle =y{\left(1-4x^2{y}^2\right)}^{-\frac{3}{2}}\cdot \left(-8x{y}^2\right)}$

${\displaystyle =-\frac{8x{y}^3}{\left(1-4x^2{y}^2\right)\sqrt{1-4x^2{y}^2}}}$

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

${\displaystyle =x{\left(1-4x^2{y}^2\right)}^{-\frac{3}{2}}\cdot \left(-8x^{2}y\right)}$

${\displaystyle =-\frac{8x^{3}y}{\left(1-4x^2{y}^2\right)\sqrt{1-4x^2{y}^2}}}$

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

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

積の微分の公式を用いて,

${\displaystyle =\frac{-2+8x^2{y}^2-8x^2{y}^2}{\left(1-4x^2{y}^2\right)\sqrt{1-4x^2{y}^2}}}$

${\displaystyle =-\frac{2}{\left(1-4x^2{y}^2\right)\sqrt{1-4x^2{y}^2}}}$　･･････(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 x\partial y}}$${\displaystyle =\frac{\partial }{\partial x}\left(\frac{\partial z}{\partial y}\right)}$

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

${\displaystyle =\frac{-2\left(1-4x^2{y}^2\right)-8x{y}^2}{\left(1-4x^2{y}^2\right)\sqrt{1-4x^2{y}^2}}}$

${\displaystyle =-\frac{2}{\left(1-4x^2{y}^2\right)\sqrt{1-4x^2{y}^2}}}$　･･････(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．

　

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31日