TITLE: Showing a property using Jacobians
QUESTION [0 upvotes]: Show that $\left(\displaystyle\frac{\partial (u,v)}{\partial (x,y)}\right)\left(\displaystyle\frac{\partial (x,y)}{\partial (r,s)}\right)=\displaystyle\frac{\partial (u,v)}{\partial (r,s)}$. Thus, prove that $\displaystyle\frac{\partial (u,v)}{\partial (x,y)}=\frac{1}{\displaystyle\frac{\partial (x,y)}{\partial (u,v)}}$ 

I have no idea on how to get started with this problem. I tried attacking it down with Jacobians, but to no avail. I'm not even sure how did they define the systems in this case.

REPLY [1 votes]: Lemma: Let  $A, B, C$ be $m \times m $ matrices such that $A \times B= C$, then $detC= detA \cdot detB$. 
Proof: By lemma, it's sufficient to show that $$ \begin{pmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x}& \frac{\partial v}{\partial y} \end{pmatrix} \begin{pmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial s}\\ \frac{\partial y}{\partial r}& \frac{\partial y}{\partial s} \end{pmatrix}= \begin{pmatrix}\frac{\partial u}{\partial r} & \frac{\partial u}{\partial s}\\ \frac{\partial v}{\partial r}& \frac{\partial v}{\partial s} \end{pmatrix}$$
However, it is directly given by the chain rule, since according to the chain rule, regarding (u,v) as the composition of (x,y) and (r,s), we have, for example
$$\frac{\partial u}{\partial r}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial r}$$ And other three can be checked as well. 
Therefore we have $$ \begin{pmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x}& \frac{\partial v}{\partial y} \end{pmatrix} \begin{pmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial s}\\ \frac{\partial y}{\partial r}& \frac{\partial y}{\partial s} \end{pmatrix}= \begin{pmatrix}\frac{\partial u}{\partial r} & \frac{\partial u}{\partial s}\\ \frac{\partial v}{\partial r}& \frac{\partial v}{\partial s} \end{pmatrix}$$
Thus $(\displaystyle\frac{\partial (u,v)}{\partial (x,y)})(\displaystyle\frac{\partial (x,y)}{\partial (r,s)})=\displaystyle\frac{\partial (u,v)}{\partial (r,s)}$
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