次の関数の2階偏導関数 $f_{xx}$, $f_{yy}$ と交差偏導関数 $f_{xy}$, $f_{yx}$ を求め、同じ値になることを確認します。 (1) $f(x, y) = xy^2 - 3xy^2$ (2) $f(x, y) = \frac{2}{3}x^3y - \frac{1}{4}xy^2$ (3) $f(x, y) = x^{\frac{1}{4}}y - x^3y^{\frac{1}{3}}$

1. 問題の内容

次の関数の2階偏導関数 $f_{xx}$, $f_{yy}$ と交差偏導関数 $f_{xy}$, $f_{yx}$ を求め、同じ値になることを確認します。

(1) $f(x, y) = xy^2 - 3xy^2$

(2) $f(x, y) = \frac{2}{3}x^3y - \frac{1}{4}xy^2$

(3) $f(x, y) = x^{\frac{1}{4}}y - x^3y^{\frac{1}{3}}$

2. 解き方の手順

偏導関数を計算する手順は以下の通りです。

(1) $f(x, y) = xy^2 - 3xy^2 = -2xy^2$

* $f_x = \frac{\partial f}{\partial x} = -2y^2$

* $f_y = \frac{\partial f}{\partial y} = -4xy$

* $f_{xx} = \frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x} (-2y^2) = 0$

* $f_{yy} = \frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y} (-4xy) = -4x$

* $f_{xy} = \frac{\partial^2 f}{\partial y \partial x} = \frac{\partial}{\partial y} (-2y^2) = -4y$

* $f_{yx} = \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial}{\partial x} (-4xy) = -4y$

(2) $f(x, y) = \frac{2}{3}x^3y - \frac{1}{4}xy^2$

* $f_x = \frac{\partial f}{\partial x} = 2x^2y - \frac{1}{4}y^2$

* $f_y = \frac{\partial f}{\partial y} = \frac{2}{3}x^3 - \frac{1}{2}xy$

* $f_{xx} = \frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x} (2x^2y - \frac{1}{4}y^2) = 4xy$

* $f_{yy} = \frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y} (\frac{2}{3}x^3 - \frac{1}{2}xy) = -\frac{1}{2}x$

* $f_{xy} = \frac{\partial^2 f}{\partial y \partial x} = \frac{\partial}{\partial y} (2x^2y - \frac{1}{4}y^2) = 2x^2 - \frac{1}{2}y$

* $f_{yx} = \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial}{\partial x} (\frac{2}{3}x^3 - \frac{1}{2}xy) = 2x^2 - \frac{1}{2}y$

(3) $f(x, y) = x^{\frac{1}{4}}y - x^3y^{\frac{1}{3}}$

* $f_x = \frac{\partial f}{\partial x} = \frac{1}{4}x^{-\frac{3}{4}}y - 3x^2y^{\frac{1}{3}}$

* $f_y = \frac{\partial f}{\partial y} = x^{\frac{1}{4}} - x^3\frac{1}{3}y^{-\frac{2}{3}} = x^{\frac{1}{4}} - \frac{1}{3}x^3y^{-\frac{2}{3}}$

* $f_{xx} = \frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x} (\frac{1}{4}x^{-\frac{3}{4}}y - 3x^2y^{\frac{1}{3}}) = -\frac{3}{16}x^{-\frac{7}{4}}y - 6xy^{\frac{1}{3}}$

* $f_{yy} = \frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y} (x^{\frac{1}{4}} - \frac{1}{3}x^3y^{-\frac{2}{3}}) = \frac{2}{9}x^3y^{-\frac{5}{3}}$

* $f_{xy} = \frac{\partial^2 f}{\partial y \partial x} = \frac{\partial}{\partial y} (\frac{1}{4}x^{-\frac{3}{4}}y - 3x^2y^{\frac{1}{3}}) = \frac{1}{4}x^{-\frac{3}{4}} - x^2y^{-\frac{2}{3}}$

* $f_{yx} = \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial}{\partial x} (x^{\frac{1}{4}} - \frac{1}{3}x^3y^{-\frac{2}{3}}) = \frac{1}{4}x^{-\frac{3}{4}} - x^2y^{-\frac{2}{3}}$

3. 最終的な答え

(1) $f(x, y) = xy^2 - 3xy^2$

* $f_{xx} = 0$

* $f_{yy} = -4x$

* $f_{xy} = -4y$

* $f_{yx} = -4y$

(2) $f(x, y) = \frac{2}{3}x^3y - \frac{1}{4}xy^2$

* $f_{xx} = 4xy$

* $f_{yy} = -\frac{1}{2}x$

* $f_{xy} = 2x^2 - \frac{1}{2}y$

* $f_{yx} = 2x^2 - \frac{1}{2}y$

(3) $f(x, y) = x^{\frac{1}{4}}y - x^3y^{\frac{1}{3}}$

* $f_{xx} = -\frac{3}{16}x^{-\frac{7}{4}}y - 6xy^{\frac{1}{3}}$

* $f_{yy} = \frac{2}{9}x^3y^{-\frac{5}{3}}$

* $f_{xy} = \frac{1}{4}x^{-\frac{3}{4}} - x^2y^{-\frac{2}{3}}$

* $f_{yx} = \frac{1}{4}x^{-\frac{3}{4}} - x^2y^{-\frac{2}{3}}$