multiplication of derivatives of rotate paras gives 1

讨论的公式为:

{(\frac{\partial z}{\partial x})}_{y} {(\frac{\partial x}{\partial y})}_{z} {(\frac{\partial y}{\partial z})}_{x} = - 1

我们可以看到这个公式好奇怪,现在尝试证明一下

方法一

我们首先有二元函数

z = z (x, y)

$z = z(x,y) = z(x, y(x,z))$ .

{(\frac{\partial z}{\partial x})}_{z} = {(\frac{\partial z}{\partial x})}_{y} + {(\frac{\partial z}{\partial y})}_{x} {(\frac{\partial y}{\partial x})}_{z}

$z$ $\left(\frac{\partial z}{\partial x}\right)$ 当然为0. 于是我们得到:

\begin{aligned} 0 & = {(\frac{\partial z}{\partial x})}_{y} + {(\frac{\partial z}{\partial y})}_{x} {(\frac{\partial y}{\partial x})}_{z} \\ \Rightarrow & - {(\frac{\partial z}{\partial x})}_{y} & = {(\frac{\partial z}{\partial y})}_{x} {(\frac{\partial y}{\partial x})}_{z} \\ \Rightarrow & - 1 & = {(\frac{\partial x}{\partial z})}_{y} {(\frac{\partial z}{\partial y})}_{x} {(\frac{\partial y}{\partial x})}_{z} \end{aligned}

这就是我们希望给出的结论.

使用同样的二元函数, 这次用如下的方法表示微分

\begin{matrix} d x = {(\frac{\partial x}{\partial y})}_{z} d y + {(\frac{\partial x}{\partial z})}_{y} d z \\ d y = {(\frac{\partial y}{\partial x})}_{z} d x + {(\frac{\partial y}{\partial z})}_{x} d z \end{matrix}

把第二式带入第一式,可以得到

d x = {(\frac{\partial x}{\partial y})}_{z} {(\frac{\partial y}{\partial x})}_{z} d x + [{(\frac{\partial x}{\partial y})}_{z} {(\frac{\partial y}{\partial z})}_{x} + {(\frac{\partial x}{\partial z})}_{y}] d z

直接对比两边,我们得到:

\begin{matrix} {(\frac{\partial x}{\partial y})}_{z} = \frac{1}{{(\frac{\partial y}{\partial x})}_{z}} \\ {(\frac{\partial x}{\partial y})}_{z} {(\frac{\partial y}{\partial z})}_{x} {(\frac{\partial z}{\partial x})}_{y} = - 1 \end{matrix}

...

好难画啊