(1) $z = xy$, $x = \sin^{-1}(uv)$, $y = \cos^{-1}(uv)$ のとき、$\frac{\partial z}{\partial u}$ と $\frac{\partial z}{\partial v}$ を求めよ。 (2) $z = \log(x^2 + y^2)$, $x = u - v$, $y = u + v$ のとき、$\frac{\partial z}{\partial u}$ と $\frac{\partial z}{\partial v}$ を求めよ。

1. 問題の内容

(1) $z = xy$, $x = \sin^{-1}(uv)$, $y = \cos^{-1}(uv)$ のとき、$\frac{\partial z}{\partial u}$ と $\frac{\partial z}{\partial v}$ を求めよ。

(2) $z = \log(x^2 + y^2)$, $x = u - v$, $y = u + v$ のとき、$\frac{\partial z}{\partial u}$ と $\frac{\partial z}{\partial v}$ を求めよ。

2. 解き方の手順

(1) 合成関数の微分公式を使う。

$\frac{\partial z}{\partial u} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial u}$

$\frac{\partial z}{\partial v} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial v} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial v}$

$\frac{\partial z}{\partial x} = y = \cos^{-1}(uv)$

$\frac{\partial z}{\partial y} = x = \sin^{-1}(uv)$

$\frac{\partial x}{\partial u} = \frac{1}{\sqrt{1 - (uv)^2}} \cdot v = \frac{v}{\sqrt{1 - u^2v^2}}$

$\frac{\partial x}{\partial v} = \frac{1}{\sqrt{1 - (uv)^2}} \cdot u = \frac{u}{\sqrt{1 - u^2v^2}}$

$\frac{\partial y}{\partial u} = \frac{-1}{\sqrt{1 - (uv)^2}} \cdot v = \frac{-v}{\sqrt{1 - u^2v^2}}$

$\frac{\partial y}{\partial v} = \frac{-1}{\sqrt{1 - (uv)^2}} \cdot u = \frac{-u}{\sqrt{1 - u^2v^2}}$

したがって、

$\frac{\partial z}{\partial u} = \cos^{-1}(uv) \cdot \frac{v}{\sqrt{1 - u^2v^2}} + \sin^{-1}(uv) \cdot \frac{-v}{\sqrt{1 - u^2v^2}} = \frac{v(\cos^{-1}(uv) - \sin^{-1}(uv))}{\sqrt{1 - u^2v^2}}$

$\frac{\partial z}{\partial v} = \cos^{-1}(uv) \cdot \frac{u}{\sqrt{1 - u^2v^2}} + \sin^{-1}(uv) \cdot \frac{-u}{\sqrt{1 - u^2v^2}} = \frac{u(\cos^{-1}(uv) - \sin^{-1}(uv))}{\sqrt{1 - u^2v^2}}$

(2) 合成関数の微分公式を使う。

$\frac{\partial z}{\partial u} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial u}$

$\frac{\partial z}{\partial v} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial v} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial v}$

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

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

$\frac{\partial x}{\partial u} = 1$

$\frac{\partial x}{\partial v} = -1$

$\frac{\partial y}{\partial u} = 1$

$\frac{\partial y}{\partial v} = 1$

したがって、

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

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

ここで$x = u - v, y = u + v$を代入する。

$\frac{\partial z}{\partial u} = \frac{2(u - v) + 2(u + v)}{(u - v)^2 + (u + v)^2} = \frac{4u}{u^2 - 2uv + v^2 + u^2 + 2uv + v^2} = \frac{4u}{2(u^2 + v^2)} = \frac{2u}{u^2 + v^2}$

$\frac{\partial z}{\partial v} = \frac{-2(u - v) + 2(u + v)}{(u - v)^2 + (u + v)^2} = \frac{4v}{u^2 - 2uv + v^2 + u^2 + 2uv + v^2} = \frac{4v}{2(u^2 + v^2)} = \frac{2v}{u^2 + v^2}$

3. 最終的な答え

(1)

$\frac{\partial z}{\partial u} = \frac{v(\cos^{-1}(uv) - \sin^{-1}(uv))}{\sqrt{1 - u^2v^2}}$

$\frac{\partial z}{\partial v} = \frac{u(\cos^{-1}(uv) - \sin^{-1}(uv))}{\sqrt{1 - u^2v^2}}$

(2)

$\frac{\partial z}{\partial u} = \frac{2u}{u^2 + v^2}$

$\frac{\partial z}{\partial v} = \frac{2v}{u^2 + v^2}$