Seien
x
,
y
,
z
,
w
{\displaystyle x,y,z,w}
x
{\displaystyle x}
x
(
y
,
z
)
{\displaystyle x(y,z)}
x
(
z
,
w
)
{\displaystyle x(z,w)}
x
(
y
,
w
)
{\displaystyle x(y,w)}
(
∂
y
∂
x
)
z
≠
0
{\displaystyle \left({\frac {\partial y}{\partial x}}\right)_{z}\neq 0}
(
∂
x
∂
y
)
z
=
1
(
∂
y
∂
x
)
z
{\displaystyle \left({\frac {\partial x}{\partial y}}\right)_{z}={\frac {1}{\left({\frac {\partial y}{\partial x}}\right)_{z}}}}
(
∂
x
∂
y
)
z
(
∂
y
∂
z
)
x
(
∂
z
∂
x
)
y
=
−
1
{\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}
(
∂
x
∂
y
)
z
=
(
∂
x
∂
y
)
w
+
(
∂
x
∂
w
)
y
(
∂
w
∂
y
)
z
{\displaystyle \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
d
x
=
(
∂
x
∂
y
)
z
d
y
+
(
∂
x
∂
z
)
y
d
z
d
y
=
(
∂
y
∂
x
)
z
d
x
+
(
∂
y
∂
z
)
x
d
z
{\displaystyle {\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
d
y
=
(
∂
y
∂
x
)
z
d
x
+
(
∂
y
∂
z
)
x
d
z
{\displaystyle \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}
d
x
{\displaystyle \mathrm {d} x}
d
x
=
(
∂
x
∂
y
)
z
d
y
+
(
∂
x
∂
z
)
y
d
z
↓
d
y
=
(
∂
y
∂
x
)
z
d
x
+
(
∂
y
∂
z
)
x
d
z
=
(
∂
x
∂
y
)
z
⋅
(
(
∂
y
∂
x
)
z
d
x
+
(
∂
y
∂
z
)
x
d
z
)
+
(
∂
x
∂
z
)
y
d
z
=
(
∂
x
∂
y
)
z
(
∂
y
∂
x
)
z
d
x
+
(
∂
x
∂
y
)
z
(
∂
y
∂
z
)
x
d
z
+
(
∂
x
∂
z
)
y
d
z
{\displaystyle {\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
=
(
(
∂
x
∂
y
)
z
(
∂
y
∂
x
)
z
−
1
)
d
x
+
(
(
∂
x
∂
y
)
z
(
∂
y
∂
z
)
x
+
(
∂
x
∂
z
)
y
)
d
z
{\displaystyle 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
d
x
{\displaystyle \mathrm {d} x}
d
z
{\displaystyle \mathrm {d} z}
(
∂
x
∂
y
)
z
(
∂
y
∂
x
)
z
−
1
=
0
{\displaystyle \left({\frac {\partial x}{\partial y}}\right)_{z}\left({\frac {\partial y}{\partial x}}\right)_{z}-1=0}
und
(
∂
x
∂
y
)
z
(
∂
y
∂
z
)
x
+
(
∂
x
∂
z
)
y
=
0
{\displaystyle \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
(
∂
x
∂
y
)
z
(
∂
y
∂
x
)
z
−
1
=
0
⇔
(
∂
x
∂
y
)
z
(
∂
y
∂
x
)
z
=
1
⇔
(
∂
x
∂
y
)
z
=
1
(
∂
y
∂
x
)
z
{\displaystyle {\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
(
∂
x
∂
y
)
z
(
∂
y
∂
z
)
x
+
(
∂
x
∂
z
)
y
=
0
{\displaystyle \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
(
∂
x
∂
y
)
z
(
∂
y
∂
z
)
x
+
(
∂
x
∂
z
)
y
=
0
⇔
(
∂
x
∂
y
)
z
(
∂
y
∂
z
)
x
=
−
(
∂
x
∂
z
)
y
↓
(
∂
x
∂
z
)
y
=
1
(
∂
z
∂
x
)
y
⇔
(
∂
x
∂
y
)
z
(
∂
y
∂
z
)
x
=
−
1
(
∂
z
∂
x
)
y
⇔
(
∂
x
∂
y
)
z
(
∂
y
∂
z
)
x
(
∂
z
∂
x
)
y
=
−
1
{\displaystyle {\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 
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/104d21870181ccbbacd5918e0e80512e937a1a3a)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b1c69118b61b05fcf21d0ca8a1601571d5ed2a46)
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dcafb9460405b63fbb27ed896b964d96d2f02320)
