2次の偏微分

■問題

次の関数の第2次偏導関数を求めよ．

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

■答

${\displaystyle \frac{{\partial }^{2}z}{\partial x^2}=12x{y}^2{e}^{2x^3{y}^2}\left(1+3x^3{y}^2\right)}$ ，${\displaystyle \frac{{\partial }^{2}z}{\partial y^2}=4x^3{e}^{2x^3{y}^2}\left(1+4x^3{y}^2\right)}$ ，${\displaystyle \frac{{\partial }^{2}z}{\partial x\partial y}=12x^{2}y{e}^{2x^3{y}^2}\left(1+2x^3{y}^2\right)}$

■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

${\displaystyle z=e^{2x^3{y}^2}}$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．

${\displaystyle \frac{\partial z}{\partial x}=\frac{\partial }{\partial x}{e}^{2x^3{y}^2}}$

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

${\displaystyle =6x^2{y}^2{e}^{2x^3{y}^2}}$　･･････(1)

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

${\displaystyle z=e^{2x^3{y}^2}}$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．

${\displaystyle \frac{\partial z}{\partial y}=\frac{\partial }{\partial y}{e}^{2x^3{y}^2}}$

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

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

${\displaystyle =\frac{\partial }{\partial x}\left(6x^2{y}^2{e}^{2x^3{y}^2}\right)}$

${\displaystyle =12x{y}^2{e}^{2x^3{y}^2}+6x^2{y}^2\cdot 6x^2{y}^2{e}^{2x^3{y}^2}}$

${\displaystyle =12x{y}^2{e}^{2x^3{y}^2}\left(1+3x^3{y}^2\right)}$

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

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

${\displaystyle =4x^3{e}^{2x^3{y}^2}+4x^{3}y\times e^{2x^3{y}^2}\frac{\partial }{\partial y}\left(2x^3{y}^2\right)}$

${\displaystyle =4x^3{e}^{2x^3{y}^2}+4x^{3}y\cdot 4x^{3}y{e}^{2x^3{y}^2}}$

${\displaystyle =4x^3{e}^{2x^3{y}^2}\left(1+4x^3{y}^2\right)}$

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

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

${\displaystyle =\frac{\partial }{\partial y}\left(6x^2{y}^2{e}^{2x^3{y}^2}\right)}$

${\displaystyle =12x^{2}y{e}^{2x^3{y}^2}+6x^2{y}^2\cdot 4x^{3}y{e}^{2x^3{y}^2}}$

${\displaystyle =12x^{2}y{e}^{2x^3{y}^2}\left(1+2x^3{y}^2\right)}$　･･････(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．

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

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

${\displaystyle =12x^{2}y{e}^{2x^3{y}^2}+4x^{3}y\cdot 6x^2{y}^2{e}^{2x^3{y}^2}}$

${\displaystyle =12x^{2}y{e}^{2x^3{y}^2}\left(1+2x^3{y}^2\right)}$　･･････(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