# Product of Jacobians

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?

## Answers and Replies

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Samy_A
Science Advisor
Homework Helper
Try to multiply the two matrices on the left, and think about the chain rule for partial derivatives.

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) ##