(1) $z = e^{xy}$ と $x = 2u + v^2$, $y = u^3v$ の合成関数の $\frac{\partial z}{\partial u}$ と $\frac{\partial z}{\partial v}$ を求めよ。 (2) $z = \log(xy)$ と $x = u^2 + v^2$, $y = 2uv$ の合成関数の $\frac{\partial z}{\partial u}$ と $\frac{\partial z}{\partial v}$ を求めよ。

1. 問題の内容

(1) $z = e^{xy}$ と $x = 2u + v^2$, $y = u^3v$ の合成関数の $\frac{\partial z}{\partial u}$ と $\frac{\partial z}{\partial v}$ を求めよ。

(2) $z = \log(xy)$ と $x = u^2 + v^2$, $y = 2uv$ の合成関数の $\frac{\partial z}{\partial u}$ と $\frac{\partial z}{\partial v}$ を求めよ。

2. 解き方の手順

(1)

まず、$\frac{\partial z}{\partial u}$ と $\frac{\partial z}{\partial v}$ を連鎖律を用いて計算する。

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

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

それぞれの偏微分を計算する。

$\frac{\partial z}{\partial x} = ye^{xy}$

$\frac{\partial z}{\partial y} = xe^{xy}$

$\frac{\partial x}{\partial u} = 2$

$\frac{\partial x}{\partial v} = 2v$

$\frac{\partial y}{\partial u} = 3u^2v$

$\frac{\partial y}{\partial v} = u^3$

これらを代入して、

$\frac{\partial z}{\partial u} = ye^{xy}(2) + xe^{xy}(3u^2v) = 2ye^{xy} + 3xu^2ve^{xy} = (2y + 3xu^2v)e^{xy}$

$\frac{\partial z}{\partial v} = ye^{xy}(2v) + xe^{xy}(u^3) = 2vye^{xy} + xu^3e^{xy} = (2vy + xu^3)e^{xy}$

$x = 2u + v^2$ と $y = u^3v$ を代入して整理する。

$\frac{\partial z}{\partial u} = (2u^3v + 3(2u+v^2)u^2v)e^{(2u+v^2)u^3v} = (2u^3v + 6u^3v + 3u^2v^3)e^{(2u^4v + u^3v^3)} = (8u^3v + 3u^2v^3)e^{(2u^4v + u^3v^3)} = u^2v(8u + 3v^2)e^{(2u^4v + u^3v^3)}$

$\frac{\partial z}{\partial v} = (2v(u^3v) + (2u + v^2)u^3)e^{(2u+v^2)u^3v} = (2u^3v^2 + 2u^4 + u^3v^2)e^{(2u^4v + u^3v^3)} = (3u^3v^2 + 2u^4)e^{(2u^4v + u^3v^3)} = u^3(3v^2 + 2u)e^{(2u^4v + u^3v^3)}$

(2)

まず、$\frac{\partial z}{\partial u}$ と $\frac{\partial z}{\partial v}$ を連鎖律を用いて計算する。

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

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

それぞれの偏微分を計算する。

$\frac{\partial z}{\partial x} = \frac{y}{xy} = \frac{1}{x}$

$\frac{\partial z}{\partial y} = \frac{x}{xy} = \frac{1}{y}$

$\frac{\partial x}{\partial u} = 2u$

$\frac{\partial x}{\partial v} = 2v$

$\frac{\partial y}{\partial u} = 2v$

$\frac{\partial y}{\partial v} = 2u$

これらを代入して、

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

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

$x = u^2 + v^2$ と $y = 2uv$ を代入して整理する。

$\frac{\partial z}{\partial u} = \frac{2u}{u^2 + v^2} + \frac{2v}{2uv} = \frac{2u}{u^2 + v^2} + \frac{1}{u} = \frac{2u^2 + u^2 + v^2}{u(u^2+v^2)} = \frac{3u^2 + v^2}{u(u^2+v^2)}$

$\frac{\partial z}{\partial v} = \frac{2v}{u^2 + v^2} + \frac{2u}{2uv} = \frac{2v}{u^2 + v^2} + \frac{1}{v} = \frac{2v^2 + u^2 + v^2}{v(u^2+v^2)} = \frac{u^2 + 3v^2}{v(u^2+v^2)}$

3. 最終的な答え

(1)

$\frac{\partial z}{\partial u} = u^2v(8u + 3v^2)e^{(2u^4v + u^3v^3)}$

$\frac{\partial z}{\partial v} = u^3(3v^2 + 2u)e^{(2u^4v + u^3v^3)}$

(2)

$\frac{\partial z}{\partial u} = \frac{3u^2 + v^2}{u(u^2+v^2)}$

$\frac{\partial z}{\partial v} = \frac{u^2 + 3v^2}{v(u^2+v^2)}$