(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Prove the chain rule for Jacobi determinants

[itex] \frac{d(f,g)}{d(u,v)} * \frac{d(u,v)}{d(x,y)}=\frac{d(f,g)}{d(x,y)}[/itex]

2. Relevant equations

Definition of Jacobi determinant

[itex] \frac{d(f,g)}{d(u,v)} = \frac{d(f,g)}{d(u,v)} = det \begin{bmatrix}

\frac{df}{du}&\frac{df}{dv} \\

\frac{dg}{du}&\frac{dg}{dv}

\end{bmatrix} [/itex]

The determinant of a matrix product of square matrices equals the product of their determinants:

det(AB)=det(A)det(B)

3. The attempt at a solution

[itex]\frac{d(f,g)}{d(u,v)} * \frac{d(u,v)}{d(x,y)} = det \begin{bmatrix}

\frac{df}{du}&\frac{df}{dv} \\

\frac{dg}{du}&\frac{dg}{dv}

\end{bmatrix} *

det \begin{bmatrix}

\frac{du}{dx}&\frac{du}{dy} \\

\frac{dv}{dx}&\frac{dv}{dy}

\end{bmatrix} = det (\begin{bmatrix}

\frac{df}{du}&\frac{df}{dv} \\

\frac{dg}{du}&\frac{dg}{dv}

\end{bmatrix} *

\begin{bmatrix}

\frac{du}{dx}&\frac{du}{dy} \\

\frac{dv}{dx}&\frac{dv}{dy}

\end{bmatrix}) = det\begin{bmatrix}

\frac{df}{du}\frac{du}{dx} + \frac{df}{dv}\frac{dv}{dx}&\frac{df}{du}\frac{du}{dy} + \frac{df}{dv}\frac{dv}{dy} \\

\frac{dg}{du}\frac{du}{dx} + \frac{dg}{dv}\frac{dv}{dx} & \frac{dg}{du}\frac{du}{dy} + \frac{dg}{dv}\frac{dv}{dy}

\end{bmatrix} = det \begin{bmatrix}

2\frac{df}{dx} & 2\frac{df}{dy} \\

2\frac{dg}{dx}&2\frac{dg}{dy}

\end{bmatrix} = 2*\frac{d(f,g)}{d(x,y)}[/itex]

which is obviously wrong but I don't see where is my error.

Is it wrong to cancel out those cross terms after multiplying matrices, but what else can I do ?

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# Prove the chain rule for Jacobi determinants

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