与えられた3つの2変数関数 $f(x, y)$ を、それぞれ $x$ と $y$ で偏微分する。 (i) $f(x, y) = \log(\cos(xy))$ (ii) $f(x, y) = \arctan(\frac{x}{y})$ (iii) $f(x, y) = x^y$

1. 問題の内容

与えられた3つの2変数関数 $f(x, y)$ を、それぞれ $x$ と $y$ で偏微分する。

(i) $f(x, y) = \log(\cos(xy))$

(ii) $f(x, y) = \arctan(\frac{x}{y})$

(iii) $f(x, y) = x^y$

2. 解き方の手順

(i) $f(x, y) = \log(\cos(xy))$

$x$ で偏微分:

$\frac{\partial f}{\partial x} = \frac{1}{\cos(xy)} \cdot (-\sin(xy)) \cdot y = -y\frac{\sin(xy)}{\cos(xy)} = -y \tan(xy)$

$y$ で偏微分:

$\frac{\partial f}{\partial y} = \frac{1}{\cos(xy)} \cdot (-\sin(xy)) \cdot x = -x\frac{\sin(xy)}{\cos(xy)} = -x \tan(xy)$

(ii) $f(x, y) = \arctan(\frac{x}{y})$

$x$ で偏微分:

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

$y$ で偏微分:

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

(iii) $f(x, y) = x^y$

$x$ で偏微分:

$\frac{\partial f}{\partial x} = y x^{y-1}$

$y$ で偏微分:

$\frac{\partial f}{\partial y} = x^y \log(x)$

3. 最終的な答え

(i) $f(x, y) = \log(\cos(xy))$

$\frac{\partial f}{\partial x} = -y \tan(xy)$

$\frac{\partial f}{\partial y} = -x \tan(xy)$

(ii) $f(x, y) = \arctan(\frac{x}{y})$

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

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

(iii) $f(x, y) = x^y$

$\frac{\partial f}{\partial x} = y x^{y-1}$

$\frac{\partial f}{\partial y} = x^y \log(x)$