与えられた4つの二変数関数 $F(x, y)$, $G(x, y)$, $H(x, y)$, $K(x, y)$ それぞれについて、$x$ と $y$ に関する偏導関数を計算する。

解析学偏微分多変数関数

2025/7/30

1. 問題の内容

与えられた4つの二変数関数

F(x, y)

G(x, y)

H(x, y)

K(x, y)

それぞれについて、

x

と

y

に関する偏導関数を計算する。

(1)

F(x, y) = \frac{1}{1 + 2x^2 + y^2}

\frac{\partial F}{\partial x} = \frac{\partial}{\partial x} (1 + 2x^2 + y^2)^{-1} = -(1 + 2x^2 + y^2)^{-2} \cdot (4x) = \frac{-4x}{(1 + 2x^2 + y^2)^2}

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

(2)

G(x, y) = \log(x + y)

\frac{\partial G}{\partial x} = \frac{\partial}{\partial x} \log(x + y) = \frac{1}{x + y} \cdot 1 = \frac{1}{x + y}

\frac{\partial G}{\partial y} = \frac{\partial}{\partial y} \log(x + y) = \frac{1}{x + y} \cdot 1 = \frac{1}{x + y}

(3)

H(x, y) = e^{x - y}

\frac{\partial H}{\partial x} = \frac{\partial}{\partial x} e^{x - y} = e^{x - y} \cdot 1 = e^{x - y}

\frac{\partial H}{\partial y} = \frac{\partial}{\partial y} e^{x - y} = e^{x - y} \cdot (-1) = -e^{x - y}

(4)

K(x, y) = \arctan(\frac{y}{x})

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

\frac{\partial K}{\partial y} = \frac{\partial}{\partial y} \arctan(\frac{y}{x}) = \frac{1}{1 + (\frac{y}{x})^2} \cdot (\frac{1}{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}

(1)

\frac{\partial F}{\partial x} = \frac{-4x}{(1 + 2x^2 + y^2)^2}

\frac{\partial F}{\partial y} = \frac{-2y}{(1 + 2x^2 + y^2)^2}

(2)

\frac{\partial G}{\partial x} = \frac{1}{x + y}

\frac{\partial G}{\partial y} = \frac{1}{x + y}

(3)

\frac{\partial H}{\partial x} = e^{x - y}

\frac{\partial H}{\partial y} = -e^{x - y}

(4)

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

\frac{\partial K}{\partial y} = \frac{x}{x^2 + y^2}