与えられた4つの関数 $z$ について、それぞれの関数を $x$ と $y$ で偏微分する問題です。 (1) $z = e^{\frac{x}{y}}$ (2) $z = \tan^{-1}\frac{y}{x}$ (3) $z = x^y + y^x$ ($x>0, y>0$) (4) $z = \frac{x^y}{y^x}$ ($x>0, y>0$)

1. 問題の内容

与えられた4つの関数 $z$ について、それぞれの関数を $x$ と $y$ で偏微分する問題です。

(1) $z = e^{\frac{x}{y}}$

(2) $z = \tan^{-1}\frac{y}{x}$

(3) $z = x^y + y^x$ ($x>0, y>0$)

(4) $z = \frac{x^y}{y^x}$ ($x>0, y>0$)

2. 解き方の手順

(1) $z = e^{\frac{x}{y}}$ の場合

* $x$ で偏微分:

$\frac{\partial z}{\partial x} = e^{\frac{x}{y}} \cdot \frac{\partial}{\partial x}(\frac{x}{y}) = e^{\frac{x}{y}} \cdot \frac{1}{y} = \frac{1}{y} e^{\frac{x}{y}}$

* $y$ で偏微分:

$\frac{\partial z}{\partial y} = e^{\frac{x}{y}} \cdot \frac{\partial}{\partial y}(\frac{x}{y}) = e^{\frac{x}{y}} \cdot (-\frac{x}{y^2}) = -\frac{x}{y^2} e^{\frac{x}{y}}$

(2) $z = \tan^{-1}\frac{y}{x}$ の場合

* $x$ で偏微分:

$\frac{\partial z}{\partial x} = \frac{1}{1+(\frac{y}{x})^2} \cdot \frac{\partial}{\partial x}(\frac{y}{x}) = \frac{1}{1+\frac{y^2}{x^2}} \cdot (-\frac{y}{x^2}) = \frac{x^2}{x^2+y^2} \cdot (-\frac{y}{x^2}) = -\frac{y}{x^2+y^2}$

* $y$ で偏微分:

$\frac{\partial z}{\partial y} = \frac{1}{1+(\frac{y}{x})^2} \cdot \frac{\partial}{\partial y}(\frac{y}{x}) = \frac{1}{1+\frac{y^2}{x^2}} \cdot \frac{1}{x} = \frac{x^2}{x^2+y^2} \cdot \frac{1}{x} = \frac{x}{x^2+y^2}$

(3) $z = x^y + y^x$ の場合

* $x$ で偏微分:

$\frac{\partial z}{\partial x} = \frac{\partial}{\partial x}(x^y) + \frac{\partial}{\partial x}(y^x) = yx^{y-1} + y^x \ln y$

* $y$ で偏微分:

$\frac{\partial z}{\partial y} = \frac{\partial}{\partial y}(x^y) + \frac{\partial}{\partial y}(y^x) = x^y \ln x + xy^{x-1}$

(4) $z = \frac{x^y}{y^x}$ の場合

* $x$ で偏微分:

$\frac{\partial z}{\partial x} = \frac{y^x \frac{\partial}{\partial x}(x^y) - x^y \frac{\partial}{\partial x}(y^x)}{(y^x)^2} = \frac{y^x (yx^{y-1}) - x^y (y^x \ln y)}{(y^x)^2} = \frac{y^{x+1} x^{y-1} - x^y y^x \ln y}{y^{2x}} = \frac{yx^{y-1} - x^y \ln y}{y^x}$

* $y$ で偏微分:

$\frac{\partial z}{\partial y} = \frac{y^x \frac{\partial}{\partial y}(x^y) - x^y \frac{\partial}{\partial y}(y^x)}{(y^x)^2} = \frac{y^x (x^y \ln x) - x^y (xy^{x-1})}{(y^x)^2} = \frac{y^x x^y \ln x - x^{y+1} y^{x-1}}{y^{2x}} = \frac{x^y \ln x - x^{y+1}y^{-1}y^{x}}{y^x}$

$\frac{x^y(\ln x) - \frac{x^{y+1}}{y}}{y^x}$

3. 最終的な答え

(1)

$\frac{\partial z}{\partial x} = \frac{1}{y} e^{\frac{x}{y}}$

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

(2)

$\frac{\partial z}{\partial x} = -\frac{y}{x^2+y^2}$

$\frac{\partial z}{\partial y} = \frac{x}{x^2+y^2}$

(3)

$\frac{\partial z}{\partial x} = yx^{y-1} + y^x \ln y$

$\frac{\partial z}{\partial y} = x^y \ln x + xy^{x-1}$

(4)

$\frac{\partial z}{\partial x} = \frac{yx^{y-1} - x^y \ln y}{y^x}$

$\frac{\partial z}{\partial y} = \frac{x^y \ln x - x^{y+1}/y}{y^x}$