Jacobian and chain rule

Let $u$, $v$ be differentiable functions of $x$, $y$ and $x$, $y$ be differentiable functions of $s$, $t$.  Then the connection

${\displaystyle \frac{\partial (u,v)}{\partial (s,t)}}={\displaystyle \frac{\partial (u,v)}{\partial (x,y)}}\cdot {\displaystyle \frac{\partial (x,y)}{\partial (s,t)}}$(1)

between the Jacobian determinants is in .

Proof.  Starting from the right hand side of (1), where one can multiply the determinants (http://planetmath.org/Determinant2) similarly as the corresponding matrices (http://planetmath.org/MatrixMultiplication), we have

Here, the last stage has been written according to the general chain rule (http://planetmath.org/ChainRuleSeveralVariables).  But thus we have arrived at the left hand side of the equation (1), which hereby has been proved.

Remark.  The rule (1) is only a visualisation of the more general one concerning the case of functions of $n$ variables.