Aufgabensammlung Mathematik: Kettenregeln für die partielle Ableitung

Seien $x,y,z,w$ Zustandsgrößen, von denen jede von jeweils zwei anderen abhängt. Für $x$ gibt es also Funktionen $x(y,z)$, $x(z,w)$ sowie $x(y,w)$. Analoges gilt für die anderen drei Zustandsgrößen. Außerdem sei $\left({\frac {\partial y}{\partial x}}\right)_{z}\neq 0$. Man beweise:

1. $\left({\frac {\partial x}{\partial y}}\right)_{z}={\frac {1}{\left({\frac {\partial y}{\partial x}}\right)_{z}}}$
2. $\left({\frac {\partial x}{\partial y}}\right)_{z}\left({\frac {\partial y}{\partial z}}\right)_{x}\left({\frac {\partial z}{\partial x}}\right)_{y}=-1$
3. $\left({\frac {\partial x}{\partial y}}\right)_{z}=\left({\frac {\partial x}{\partial y}}\right)_{w}+\left({\frac {\partial x}{\partial w}}\right)_{y}\left({\frac {\partial w}{\partial y}}\right)_{z}$

Lösung zur 1. Teilaufgabe

Es ist

${\begin{aligned}\mathrm {d} x&=\left({\frac {\partial x}{\partial y}}\right)_{z}\mathrm {d} y+\left({\frac {\partial x}{\partial z}}\right)_{y}\mathrm {d} z\\\mathrm {d} y&=\left({\frac {\partial y}{\partial x}}\right)_{z}\mathrm {d} x+\left({\frac {\partial y}{\partial z}}\right)_{x}\mathrm {d} z\end{aligned}}$

Nun kann die Gleichung $\mathrm {d} y=\left({\frac {\partial y}{\partial x}}\right)_{z}\mathrm {d} x+\left({\frac {\partial y}{\partial z}}\right)_{x}\mathrm {d} z$ in die Gleichung für $\mathrm {d} x$ eingesetzt werden. So erhält man

${\begin{aligned}\mathrm {d} x&=\left({\frac {\partial x}{\partial y}}\right)_{z}\mathrm {d} y+\left({\frac {\partial x}{\partial z}}\right)_{y}\mathrm {d} z\\[5px]&\quad \left\downarrow \ \mathrm {d} y=\left({\frac {\partial y}{\partial x}}\right)_{z}\mathrm {d} x+\left({\frac {\partial y}{\partial z}}\right)_{x}\mathrm {d} z\right.\\[5px]&=\left({\frac {\partial x}{\partial y}}\right)_{z}\cdot \left(\left({\frac {\partial y}{\partial x}}\right)_{z}\mathrm {d} x+\left({\frac {\partial y}{\partial z}}\right)_{x}\mathrm {d} z\right)+\left({\frac {\partial x}{\partial z}}\right)_{y}\mathrm {d} z\\&=\left({\frac {\partial x}{\partial y}}\right)_{z}\left({\frac {\partial y}{\partial x}}\right)_{z}\mathrm {d} x+\left({\frac {\partial x}{\partial y}}\right)_{z}\left({\frac {\partial y}{\partial z}}\right)_{x}\mathrm {d} z+\left({\frac {\partial x}{\partial z}}\right)_{y}\mathrm {d} z\end{aligned}}$

Also ist

$0=\left(\left({\frac {\partial x}{\partial y}}\right)_{z}\left({\frac {\partial y}{\partial x}}\right)_{z}-1\right)\mathrm {d} x+\left(\left({\frac {\partial x}{\partial y}}\right)_{z}\left({\frac {\partial y}{\partial z}}\right)_{x}+\left({\frac {\partial x}{\partial z}}\right)_{y}\right)\mathrm {d} z$

Da $\mathrm {d} x$ und $\mathrm {d} z$ linear unabhängig sind, muss gelten:

$\left({\frac {\partial x}{\partial y}}\right)_{z}\left({\frac {\partial y}{\partial x}}\right)_{z}-1=0$

und

$\left({\frac {\partial x}{\partial y}}\right)_{z}\left({\frac {\partial y}{\partial z}}\right)_{x}+\left({\frac {\partial x}{\partial z}}\right)_{y}=0$

Aus der ersten Gleichung folgt

${\begin{aligned}{\begin{array}{lrl}&\left({\frac {\partial x}{\partial y}}\right)_{z}\left({\frac {\partial y}{\partial x}}\right)_{z}-1&=0\\\Leftrightarrow \ &\left({\frac {\partial x}{\partial y}}\right)_{z}\left({\frac {\partial y}{\partial x}}\right)_{z}&=1\\\Leftrightarrow \ &\left({\frac {\partial x}{\partial y}}\right)_{z}&={\frac {1}{\left({\frac {\partial y}{\partial x}}\right)_{z}}}\end{array}}\end{aligned}}$

Lösung zur 2. Teilaufgabe

In der Lösung zu zweiten Teilaufgabe wurde gezeigt, dass

$\left({\frac {\partial x}{\partial y}}\right)_{z}\left({\frac {\partial y}{\partial z}}\right)_{x}+\left({\frac {\partial x}{\partial z}}\right)_{y}=0$

Daraus folgt, dass

${\begin{aligned}{\begin{array}{lrl}&\left({\frac {\partial x}{\partial y}}\right)_{z}\left({\frac {\partial y}{\partial z}}\right)_{x}+\left({\frac {\partial x}{\partial z}}\right)_{y}&=0\\\Leftrightarrow \ &\left({\frac {\partial x}{\partial y}}\right)_{z}\left({\frac {\partial y}{\partial z}}\right)_{x}&=-\left({\frac {\partial x}{\partial z}}\right)_{y}\\[5px]&&\left\downarrow \ \left({\frac {\partial x}{\partial z}}\right)_{y}={\frac {1}{\left({\frac {\partial z}{\partial x}}\right)_{y}}}\right.\\[5px]\Leftrightarrow \ &\left({\frac {\partial x}{\partial y}}\right)_{z}\left({\frac {\partial y}{\partial z}}\right)_{x}&=-{\frac {1}{\left({\frac {\partial z}{\partial x}}\right)_{y}}}\\\Leftrightarrow \ &\left({\frac {\partial x}{\partial y}}\right)_{z}\left({\frac {\partial y}{\partial z}}\right)_{x}\left({\frac {\partial z}{\partial x}}\right)_{y}&=-1\\\end{array}}\end{aligned}}$

Lösung zur 3. Teilaufgabe

Es ist

${\begin{aligned}\mathrm {d} x&=\left({\frac {\partial x}{\partial y}}\right)_{z}\mathrm {d} y+\left({\frac {\partial x}{\partial z}}\right)_{y}\mathrm {d} z\\\mathrm {d} x&=\left({\frac {\partial x}{\partial y}}\right)_{w}\mathrm {d} y+\left({\frac {\partial x}{\partial w}}\right)_{y}\mathrm {d} w\\\mathrm {d} w&=\left({\frac {\partial w}{\partial y}}\right)_{z}\mathrm {d} y+\left({\frac {\partial w}{\partial z}}\right)_{y}\mathrm {d} z\end{aligned}}$

Wenn man nun die ersten beiden Gleichungen gleichsetzt und in dieser Gleichung $\mathrm {d} w$ durch den Term $\left({\frac {\partial w}{\partial y}}\right)_{z}\mathrm {d} y+\left({\frac {\partial w}{\partial z}}\right)_{y}\mathrm {d} z$ ersetzt, dann erhält man

${\begin{aligned}{\begin{array}{rrl}&\left({\frac {\partial x}{\partial y}}\right)_{z}\mathrm {d} y+\left({\frac {\partial x}{\partial z}}\right)_{y}\mathrm {d} z&=\left({\frac {\partial x}{\partial y}}\right)_{w}\mathrm {d} y+\left({\frac {\partial x}{\partial w}}\right)_{y}\mathrm {d} w\\[5px]&&\left\downarrow \ dw=\left({\frac {\partial w}{\partial y}}\right)_{z}\mathrm {d} y+\left({\frac {\partial w}{\partial z}}\right)_{y}\mathrm {d} z\right.\\[5px]\Rightarrow \ &\left({\frac {\partial x}{\partial y}}\right)_{z}\mathrm {d} y+\left({\frac {\partial x}{\partial z}}\right)_{y}\mathrm {d} z&=\left({\frac {\partial x}{\partial y}}\right)_{w}\mathrm {d} y+\left({\frac {\partial x}{\partial w}}\right)_{y}\cdot \left(\left({\frac {\partial w}{\partial y}}\right)_{z}\mathrm {d} y+\left({\frac {\partial w}{\partial z}}\right)_{y}\mathrm {d} z\right)\\\Rightarrow \ &\left({\frac {\partial x}{\partial y}}\right)_{z}\mathrm {d} y+\left({\frac {\partial x}{\partial z}}\right)_{y}\mathrm {d} z&=\left({\frac {\partial x}{\partial y}}\right)_{w}\mathrm {d} y+\left({\frac {\partial x}{\partial w}}\right)_{y}\left({\frac {\partial w}{\partial y}}\right)_{z}\mathrm {d} y+\left({\frac {\partial x}{\partial w}}\right)_{y}\left({\frac {\partial w}{\partial z}}\right)_{y}\mathrm {d} z\\\Rightarrow \ &\left({\frac {\partial x}{\partial y}}\right)_{z}\mathrm {d} y+\left({\frac {\partial x}{\partial z}}\right)_{y}\mathrm {d} z&=\left(\left({\frac {\partial x}{\partial y}}\right)_{w}+\left({\frac {\partial x}{\partial w}}\right)_{y}\left({\frac {\partial w}{\partial y}}\right)_{z}\right)\mathrm {d} y+\left({\frac {\partial x}{\partial w}}\right)_{y}\left({\frac {\partial w}{\partial z}}\right)_{y}\mathrm {d} z\\\Rightarrow \ &0&=\left(\left({\frac {\partial x}{\partial y}}\right)_{w}+\left({\frac {\partial x}{\partial w}}\right)_{y}\left({\frac {\partial w}{\partial y}}\right)_{z}-\left({\frac {\partial x}{\partial y}}\right)_{z}\right)\mathrm {d} y+\left(\left({\frac {\partial x}{\partial w}}\right)_{y}\left({\frac {\partial w}{\partial z}}\right)_{y}-\left({\frac {\partial x}{\partial z}}\right)_{y}\right)\mathrm {d} z\\\end{array}}\end{aligned}}$

Da $\mathrm {d} y$ und $\mathrm {d} z$ linear unabhängig sind, ist

$\left({\frac {\partial x}{\partial y}}\right)_{w}+\left({\frac {\partial x}{\partial w}}\right)_{y}\left({\frac {\partial w}{\partial y}}\right)_{z}-\left({\frac {\partial x}{\partial y}}\right)_{z}=0$

und damit

$\left({\frac {\partial x}{\partial y}}\right)_{z}=\left({\frac {\partial x}{\partial y}}\right)_{w}+\left({\frac {\partial x}{\partial w}}\right)_{y}\left({\frac {\partial w}{\partial y}}\right)_{z}$