与えられた関数 $f(x,y)$ について、2階の偏導関数 $f_{xx}$, $f_{yy}$, $f_{xy}$, $f_{yx}$ をすべて求める。

1. 問題の内容

与えられた関数 $f(x,y)$ について、2階の偏導関数 $f_{xx}$, $f_{yy}$, $f_{xy}$, $f_{yx}$ をすべて求める。

2. 解き方の手順

それぞれの関数について、以下の手順で解く。

a) $f(x,y) = x^2 y^3$

- $f_x = \frac{\partial f}{\partial x} = 2xy^3$

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

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

- $f_{yy} = \frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y} (3x^2y^2) = 6x^2y$

- $f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial}{\partial x} (3x^2y^2) = 6xy^2$

- $f_{yx} = \frac{\partial^2 f}{\partial y \partial x} = \frac{\partial}{\partial y} (2xy^3) = 6xy^2$

b) $f(x,y) = \sqrt{x^2 + y^2} = (x^2 + y^2)^{\frac{1}{2}}$

- $f_x = \frac{\partial f}{\partial x} = \frac{1}{2}(x^2 + y^2)^{-\frac{1}{2}}(2x) = \frac{x}{\sqrt{x^2 + y^2}}$

- $f_y = \frac{\partial f}{\partial y} = \frac{1}{2}(x^2 + y^2)^{-\frac{1}{2}}(2y) = \frac{y}{\sqrt{x^2 + y^2}}$

- $f_{xx} = \frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x} (\frac{x}{\sqrt{x^2 + y^2}}) = \frac{\sqrt{x^2 + y^2} - x \frac{x}{\sqrt{x^2 + y^2}}}{x^2 + y^2} = \frac{x^2 + y^2 - x^2}{(x^2 + y^2)^{\frac{3}{2}}} = \frac{y^2}{(x^2 + y^2)^{\frac{3}{2}}}$

- $f_{yy} = \frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y} (\frac{y}{\sqrt{x^2 + y^2}}) = \frac{\sqrt{x^2 + y^2} - y \frac{y}{\sqrt{x^2 + y^2}}}{x^2 + y^2} = \frac{x^2 + y^2 - y^2}{(x^2 + y^2)^{\frac{3}{2}}} = \frac{x^2}{(x^2 + y^2)^{\frac{3}{2}}}$

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

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

c) $f(x,y) = 2x \ln(x^2 + y^2)$

- $f_x = 2 \ln(x^2 + y^2) + 2x \frac{2x}{x^2 + y^2} = 2 \ln(x^2 + y^2) + \frac{4x^2}{x^2 + y^2}$

- $f_y = 2x \frac{2y}{x^2 + y^2} = \frac{4xy}{x^2 + y^2}$

- $f_{xx} = \frac{2(2x)}{x^2 + y^2} + \frac{8x(x^2 + y^2) - 4x^2(2x)}{(x^2 + y^2)^2} = \frac{4x}{x^2 + y^2} + \frac{8x^3 + 8xy^2 - 8x^3}{(x^2 + y^2)^2} = \frac{4x}{x^2 + y^2} + \frac{8xy^2}{(x^2 + y^2)^2} = \frac{4x(x^2 + y^2) + 8xy^2}{(x^2 + y^2)^2} = \frac{4x^3 + 12xy^2}{(x^2 + y^2)^2}$

- $f_{yy} = \frac{4x(x^2 + y^2) - 4xy(2y)}{(x^2 + y^2)^2} = \frac{4x^3 + 4xy^2 - 8xy^2}{(x^2 + y^2)^2} = \frac{4x^3 - 4xy^2}{(x^2 + y^2)^2}$

- $f_{xy} = \frac{4y(x^2 + y^2) - 4xy(2x)}{(x^2 + y^2)^2} = \frac{4x^2y + 4y^3 - 8x^2y}{(x^2 + y^2)^2} = \frac{4y^3 - 4x^2y}{(x^2 + y^2)^2}$

- $f_{yx} = \frac{4y(x^2 + y^2) - 4xy(2x)}{(x^2 + y^2)^2} = \frac{4x^2y + 4y^3 - 8x^2y}{(x^2 + y^2)^2} = \frac{4y^3 - 4x^2y}{(x^2 + y^2)^2}$

d) $f(x,y) = e^{x^2 + y^2}$

- $f_x = 2xe^{x^2 + y^2}$

- $f_y = 2ye^{x^2 + y^2}$

- $f_{xx} = 2e^{x^2 + y^2} + 2x(2xe^{x^2 + y^2}) = 2e^{x^2 + y^2} + 4x^2e^{x^2 + y^2} = (2 + 4x^2)e^{x^2 + y^2}$

- $f_{yy} = 2e^{x^2 + y^2} + 2y(2ye^{x^2 + y^2}) = 2e^{x^2 + y^2} + 4y^2e^{x^2 + y^2} = (2 + 4y^2)e^{x^2 + y^2}$

- $f_{xy} = 2x(2ye^{x^2 + y^2}) = 4xye^{x^2 + y^2}$

- $f_{yx} = 2y(2xe^{x^2 + y^2}) = 4xye^{x^2 + y^2}$

3. 最終的な答え

a) $f(x,y) = x^2 y^3$

- $f_{xx} = 2y^3$

- $f_{yy} = 6x^2y$

- $f_{xy} = 6xy^2$

- $f_{yx} = 6xy^2$

b) $f(x,y) = \sqrt{x^2 + y^2}$

- $f_{xx} = \frac{y^2}{(x^2 + y^2)^{\frac{3}{2}}}$

- $f_{yy} = \frac{x^2}{(x^2 + y^2)^{\frac{3}{2}}}$

- $f_{xy} = \frac{-xy}{(x^2 + y^2)^{\frac{3}{2}}}$

- $f_{yx} = \frac{-xy}{(x^2 + y^2)^{\frac{3}{2}}}$

c) $f(x,y) = 2x \ln(x^2 + y^2)$

- $f_{xx} = \frac{4x^3 + 12xy^2}{(x^2 + y^2)^2}$

- $f_{yy} = \frac{4x^3 - 4xy^2}{(x^2 + y^2)^2}$

- $f_{xy} = \frac{4y^3 - 4x^2y}{(x^2 + y^2)^2}$

- $f_{yx} = \frac{4y^3 - 4x^2y}{(x^2 + y^2)^2}$

d) $f(x,y) = e^{x^2 + y^2}$

- $f_{xx} = (2 + 4x^2)e^{x^2 + y^2}$

- $f_{yy} = (2 + 4y^2)e^{x^2 + y^2}$

- $f_{xy} = 4xye^{x^2 + y^2}$

- $f_{yx} = 4xye^{x^2 + y^2}$