2次の偏微分

■問題

次の関数の第2次偏導関数を求めよ．

${\displaystyle z=\cos x^2{y}^2}$

■答

${\displaystyle \frac{{\partial }^{2}z}{\partial x^2}=-2y^2\sin x^{2}y{}^2-4x^2{y}^4\cos x^{2}y{}^2}$

${\displaystyle \frac{{\partial }^{2}z}{\partial y^2}=-2x^2\sin x^{2}y{}^2-4x^4{y}^2\cos x^{2}y{}^2}$

${\displaystyle \frac{{\partial }^{2}z}{\partial y\partial x}=\frac{{\partial }^{2}z}{\partial x\partial y}}$ ${\displaystyle =-4xy\sin x^2{y}^2-4x^3{y}^3\cos x^2{y}^2}$

■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

${\displaystyle z=\cos x^2{y}^2}$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．

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

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

${\displaystyle z=\cos x^2{y}^2}$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．

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

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

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

${\displaystyle =-2y^2\sin x^{2}y{}^2-2x{y}^2\cdot 2x{y}^2\cos x^{2}y{}^2}$

${\displaystyle =-2y^2\sin x^{2}y{}^2-4x^2{y}^4\cos x^{2}y{}^2}$

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

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

${\displaystyle =-2x^2\sin x^{2}y{}^2-2x^{2}y\cdot 2x^{2}y\cos x^{2}y{}^2}$

${\displaystyle =-2x^2\sin x^{2}y{}^2-4x^4{y}^2\cos x^{2}y{}^2}$

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

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

積の微分の公式を用いて

${\displaystyle =-4xy\sin x^2{y}^2-2x{y}^2\cdot 2x^{2}y\cos x^2{y}^2}$

${\displaystyle =-4xy\sin x^2{y}^2-4x^3{y}^3\cos x^2{y}^2}$　･･････(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．

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

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

${\displaystyle =-4xy\sin x^2{y}^2-2x^{2}y\cos x^2{y}^2\times 2x{y}^2}$

${\displaystyle =-4xy\sin x^2{y}^2-4x^3{y}^3\cos x^2{y}^2}$　･･････(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