$x, y, z$ の間に関数関係 $f(x, y, z) = 0$ があるとき、次の式が成り立つことを示す問題です。 $$ \left( \frac{\partial x}{\partial y} \right)_z \left( \frac{\partial y}{\partial z} \right)_x \left( \frac{\partial z}{\partial x} \right)_y = -1 $$

1. 問題の内容

$x, y, z$ の間に関数関係 $f(x, y, z) = 0$ があるとき、次の式が成り立つことを示す問題です。

$$\left( \frac{\partial x}{\partial y} \right)_z \left( \frac{\partial y}{\partial z} \right)_x \left( \frac{\partial z}{\partial x} \right)_y = -1$$

2. 解き方の手順

$f(x, y, z) = 0$ の全微分を考えます。

$$df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy + \frac{\partial f}{\partial z} dz = 0$$

($\frac{\partial x}{\partial y})_z$ を求めるには、$z$を固定して、$x$と$y$の変化を考えます。$dz=0$とおくと、

$$\frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy = 0$$

$$\left( \frac{\partial x}{\partial y} \right)_z = - \frac{\frac{\partial f}{\partial y}}{\frac{\partial f}{\partial x}}$$

同様に、

$$\left( \frac{\partial y}{\partial z} \right)_x = - \frac{\frac{\partial f}{\partial z}}{\frac{\partial f}{\partial y}}$$

$$\left( \frac{\partial z}{\partial x} \right)_y = - \frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial z}}$$

したがって、

$$\left( \frac{\partial x}{\partial y} \right)_z \left( \frac{\partial y}{\partial z} \right)_x \left( \frac{\partial z}{\partial x} \right)_y = \left( - \frac{\frac{\partial f}{\partial y}}{\frac{\partial f}{\partial x}} \right) \left( - \frac{\frac{\partial f}{\partial z}}{\frac{\partial f}{\partial y}} \right) \left( - \frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial z}} \right) = -1$$

3. 最終的な答え

$$\left( \frac{\partial x}{\partial y} \right)_z \left( \frac{\partial y}{\partial z} \right)_x \left( \frac{\partial z}{\partial x} \right)_y = -1$$