# Product of Jacobians

1. Dec 7, 2015

### Alex Mamaev

Question:
The following are determinants of partial derivatives multiplied together giving another determinant of partial derivatives
Prove that this equality holds:

Relevant Equations:
|du/dx du/dy| |dx/dr dx/ds| |du/dr du/ds|
|dv/dx dv/dy| |dy/dr dy/ds| = |dv/dr dv/ds|

Attempt at Solution:
I took the determinant of first and second matrices, multiplied them together however the answer I got was twice the determinant that I expected. [ I expected to get: (du/dr)(dv/ds)-(du/ds)(dv/dr)]. Then when i tried to use the rule that the product of determinants is the determinant of products and took the determinant of the multiplied matrices i got four times the determinant that i expected.

Does this constant multiple matter or did I make a mistake. Pretty sure i got the algebra correct so is there an error in my reasoning?

2. Dec 7, 2015

### Samy_A

Try to multiply the two matrices on the left, and think about the chain rule for partial derivatives.

3. Dec 7, 2015

### geoffrey159

Try to apply the composition formula for differentials, which says :
$f$ differentiable in $a$ and $g$ differentiable in $f(a)$ $\Rightarrow$ $g\circ f$ differentiable in $a$ and $d_a(g\circ f) = d_{f(a)}(g) \circ d_a(f)$

This translates in term of Jacobians by: $J_{g\circ f}(a) = J_g(f(a)) \times J_f(a)$

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted