偏微分の関係式の導出

問題

以下の関係式を導出せよ。

(1) $\displaystyle \left( \frac{\partial Z}{\partial X} \right)_Y = \left( \frac{\partial X}{\partial Z} \right)^{-1}_Y$

(2) $\displaystyle \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 $

解答

(1)
$X(Y, Z)$として、全微分すると

\begin{eqnarray*}
\diff X =\left( \frac{\partial X}{\partial Y} \right)_Z \diff Y + \left( \frac{\partial X}{\partial Z} \right)_Y \diff Z
\end{eqnarray*}

と表される。

一方、$Z(X, Y)$として、全微分すると

\begin{eqnarray*}
\diff Z =\left( \frac{\partial Z}{\partial X} \right)_Y \diff X + \left( \frac{\partial Z}{\partial Y} \right)_X \diff Y
\end{eqnarray*}

と表される。

2式より、

\begin{eqnarray*}
\diff Z &=& \left( \frac{\partial Z}{\partial X} \right)_Y \diff X + \left( \frac{\partial Z}{\partial Y} \right)_X \diff Y \\
\\
&=& \left( \frac{\partial Z}{\partial X} \right)_Y \left[ \left( \frac{\partial X}{\partial Y} \right)_Z \diff Y + \left( \frac{\partial X}{\partial Z} \right)_Y \diff Z \right] + \left( \frac{\partial Z}{\partial Y} \right)_X \diff Y \\
\\
&=&
\left( \frac{\partial Z}{\partial X} \right)_Y \left( \frac{\partial X}{\partial Y} \right)_Z \diff Y +
\left( \frac{\partial Z}{\partial X} \right)_Y \left( \frac{\partial X}{\partial Z} \right)_Y \diff Z +
\left( \frac{\partial Z}{\partial Y} \right)_X \diff Y \\
\\
&=&
\left[
\left( \frac{\partial Z}{\partial X} \right)_Y \left( \frac{\partial X}{\partial Y} \right)_Z +
\left( \frac{\partial Z}{\partial Y} \right)_X
+ \right] \diff Y +
\left( \frac{\partial Z}{\partial X} \right)_Y \left( \frac{\partial X}{\partial Z} \right)_Y \diff Z \\
\\ \\
0 &=& \left[
\left( \frac{\partial Z}{\partial X} \right)_Y \left( \frac{\partial X}{\partial Y} \right)_Z +
\left( \frac{\partial Z}{\partial Y} \right)_X
+ \right] \diff Y +
\left[
\left( \frac{\partial Z}{\partial X} \right)_Y \left( \frac{\partial X}{\partial Z} \right)_Y -1 \right] \diff Z
\end{eqnarray*}

となる。
従って、
\begin{eqnarray*}
\left( \frac{\partial Z}{\partial X} \right)_Y \left( \frac{\partial X}{\partial Y} \right)_Z +
\left( \frac{\partial Z}{\partial Y} \right)_X &=& 0 \\
\\
\left( \frac{\partial Z}{\partial X} \right)_Y \left( \frac{\partial X}{\partial Z} \right)_Y -1 &=& 0
\end{eqnarray*}
であるから

(1)については
\begin{eqnarray*}
\left( \frac{\partial Z}{\partial X} \right)_Y \left( \frac{\partial X}{\partial Z} \right)_Y -1 &=& 0 \\
\\
\left( \frac{\partial Z}{\partial X} \right)_Y \left( \frac{\partial X}{\partial Z} \right)_Y &=& 1 \\
\\
\left( \frac{\partial Z}{\partial X} \right)_Y &=& \left( \frac{\partial X}{\partial Z} \right)^{-1}_Y
\end{eqnarray*}
となる。

(2)については
\begin{eqnarray*}
\left( \frac{\partial Z}{\partial X} \right)_Y \left( \frac{\partial X}{\partial Y} \right)_Z +
\left( \frac{\partial Z}{\partial Y} \right)_X &=& 0 \\
\\
\left( \frac{\partial Z}{\partial X} \right)_Y \left( \frac{\partial X}{\partial Y} \right)_Z &=& –
\left( \frac{\partial Z}{\partial Y} \right)_X \\
\\
\left( \frac{\partial Z}{\partial X} \right)_Y \left( \frac{\partial X}{\partial Y} \right)_Z \left( \frac{\partial Y}{\partial Z} \right)_X &=& –
\left( \frac{\partial Z}{\partial Y} \right)_X \left( \frac{\partial Y}{\partial Z} \right)_X \\
\\
\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
\end{eqnarray*}
となる。