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Product of Jacobians

  1. Dec 7, 2015 #1
    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. jcsd
  3. Dec 7, 2015 #2

    Samy_A

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    Science Advisor
    Homework Helper

    Try to multiply the two matrices on the left, and think about the chain rule for partial derivatives.
     
  4. Dec 7, 2015 #3
    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) ##
     
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