関数 $f(x,y) = \frac{\sqrt{x+y}}{xy}$ の、xとyに関する1階および2階の偏導関数を求めます。つまり、$f_x$, $f_y$, $f_{xx}$, $f_{yy}$, $f_{xy}$, $f_{yx}$ を計算します。

はい、承知いたしました。画像にある数学の問題を解いていきます。今回は、問題 (3) $f(x,y) = \frac{\sqrt{x+y}}{xy}$ の2階までの偏導関数を求めます。

1. 問題の内容

関数 $f(x,y) = \frac{\sqrt{x+y}}{xy}$ の、xとyに関する1階および2階の偏導関数を求めます。つまり、$f_x$, $f_y$, $f_{xx}$, $f_{yy}$, $f_{xy}$, $f_{yx}$ を計算します。

2. 解き方の手順

まず、関数を変形します。

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

次に、1階偏導関数を求めます。

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

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

次に、2階偏導関数を求めます。

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

$f_{xx} = \frac{\partial}{\partial x}\left(\frac{-x-2y}{2yx^2\sqrt{x+y}} \right) = -\frac{1}{2y}\frac{\partial}{\partial x}\left(\frac{x+2y}{x^2\sqrt{x+y}} \right)$

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

$f_{xx}= -\frac{1}{2y} \frac{x^2(x+y) - (x+2y) \left(2x(x+y) + \frac{x^2}{2} \right)}{x^4 (x+y)^{3/2}} = \frac{3x^2 + 8xy + 8y^2}{4x^3 y (x+y)^{3/2}}$

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

$f_{yy}= \frac{\partial}{\partial y}\left(\frac{-2x-y}{2xy^2\sqrt{x+y}} \right) = -\frac{1}{2x}\frac{\partial}{\partial y}\left(\frac{2x+y}{y^2\sqrt{x+y}} \right)$

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

$f_{yy} = -\frac{1}{2x} \frac{y^2(x+y) - (2x+y) \left(2y(x+y) + \frac{y^2}{2} \right)}{y^4 (x+y)^{3/2}} = \frac{8x^2 + 8xy + 3y^2}{4xy^3(x+y)^{3/2}}$

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

$f_{xy} = -\frac{1}{2y^2} \frac{(2x(x+y)^{1/2}) - (2x+y)\left((x+y)^{1/2} + \frac{x}{2(x+y)^{1/2}}\right)}{x^2(x+y)}$

$f_{xy} = -\frac{1}{2y^2} \frac{2x(x+y) - (2x+y)\left((x+y) + \frac{x}{2}\right)}{x^2(x+y)^{3/2}} = \frac{4x^2 + 4xy + 3xy + y^2}{4x^2 y^2 (x+y)^{3/2}} = \frac{4x^2 + 7xy + y^2}{4x^2 y^2 (x+y)^{3/2}}$

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

$f_{yx} = -\frac{1}{2x^2} \frac{(2y(x+y)^{1/2}) - (x+2y)\left((x+y)^{1/2} + \frac{y}{2(x+y)^{1/2}}\right)}{y^2(x+y)}$

$f_{yx} = -\frac{1}{2x^2} \frac{2y(x+y) - (x+2y)\left((x+y) + \frac{y}{2}\right)}{y^2(x+y)^{3/2}} = \frac{x^2 + 7xy + 4y^2}{4 x^2y^2 (x+y)^{3/2}}$

3. 最終的な答え

$f_x = \frac{-x-2y}{2x^2y\sqrt{x+y}}$

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

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

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

$f_{xy} = \frac{4x^2 + 7xy + y^2}{4x^2 y^2 (x+y)^{3/2}}$

$f_{yx} = \frac{x^2 + 7xy + 4y^2}{4 x^2y^2 (x+y)^{3/2}}$