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)}\;=\;\frac{\partial(u,v)}{% \partial(x,y)}\cdot\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

\left|\begin{matrix}u_{x}&u_{y}\\ v_{x}&v_{y}\\ \end{matrix}\right|\cdot\left|\begin{matrix}x_{s}&x_{t}\\ y_{s}&y_{t}\\ \end{matrix}\right|\;=\;\left|\begin{matrix}u_{x}x_{s}+u_{y}y_{s}&u_{x}x_{t}+u% _{y}y_{t}\\ v_{x}x_{s}+v_{y}y_{s}&v_{x}x_{t}+v_{y}y_{t}\\ \end{matrix}\right|\;=\;\left|\begin{matrix}u_{s}&u_{t}\\ v_{s}&v_{t}\\ \end{matrix}\right|.

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.

Title	Jacobian and chain rule
Canonical name	JacobianAndChainRule
Date of creation	2013-03-22 18:59:45
Last modified on	2013-03-22 18:59:45
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	4
Author	pahio (2872)
Entry type	Example
Classification	msc 15-00
Classification	msc 26B05
Classification	msc 26B10