# Homework Help: Prove the chain rule for Jacobi determinants

1. Jan 10, 2012

### vst98

1. The problem statement, all variables and given/known data

Prove the chain rule for Jacobi determinants

$\frac{d(f,g)}{d(u,v)} * \frac{d(u,v)}{d(x,y)}=\frac{d(f,g)}{d(x,y)}$

2. Relevant equations

Definition of Jacobi determinant

$\frac{d(f,g)}{d(u,v)} = \frac{d(f,g)}{d(u,v)} = det \begin{bmatrix} \frac{df}{du}&\frac{df}{dv} \\ \frac{dg}{du}&\frac{dg}{dv} \end{bmatrix}$

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

$\frac{d(f,g)}{d(u,v)} * \frac{d(u,v)}{d(x,y)} = det \begin{bmatrix} \frac{df}{du}&\frac{df}{dv} \\ \frac{dg}{du}&\frac{dg}{dv} \end{bmatrix} * det \begin{bmatrix} \frac{du}{dx}&\frac{du}{dy} \\ \frac{dv}{dx}&\frac{dv}{dy} \end{bmatrix} = det (\begin{bmatrix} \frac{df}{du}&\frac{df}{dv} \\ \frac{dg}{du}&\frac{dg}{dv} \end{bmatrix} * \begin{bmatrix} \frac{du}{dx}&\frac{du}{dy} \\ \frac{dv}{dx}&\frac{dv}{dy} \end{bmatrix}) = det\begin{bmatrix} \frac{df}{du}\frac{du}{dx} + \frac{df}{dv}\frac{dv}{dx}&\frac{df}{du}\frac{du}{dy} + \frac{df}{dv}\frac{dv}{dy} \\ \frac{dg}{du}\frac{du}{dx} + \frac{dg}{dv}\frac{dv}{dx} & \frac{dg}{du}\frac{du}{dy} + \frac{dg}{dv}\frac{dv}{dy} \end{bmatrix} = det \begin{bmatrix} 2\frac{df}{dx} & 2\frac{df}{dy} \\ 2\frac{dg}{dx}&2\frac{dg}{dy} \end{bmatrix} = 2*\frac{d(f,g)}{d(x,y)}$

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 ?

2. Nov 6, 2013

### aalpie

Proof

I know this is an old thread (and I hope you are not still stuck:tongue:), but I am just replying so other can use it for future reference.

The problem lies in the fact that you see the jacobian matrix as a matrix containing derivatives. As it is related to a multivariate function (a function depending on multiple variables), you should work with partial derivatives. As f is a function of u and v you can write

$df=\left(\frac{\partial f}{\partial u}\right)_{v}du+\left(\frac{\partial f}{\partial v}\right)_{u}dv$

where the subscript $\left(\frac{\partial f}{\partial u}\right)_{x}$ means the partial derivative of f with respect to u holding x constant.

As u and v are a function of x and y you can write:
$du=\left(\frac{\partial u}{\partial x}\right)_{y}dx+\left(\frac{\partial u}{\partial y}\right)_{x}dy$
$dv=\left(\frac{\partial v}{\partial x}\right)_{y}dx+\left(\frac{\partial v}{\partial y}\right)_{x}dy$

If we will in these equations in the previous, we can obtain:
$df=\left[\left(\frac{\partial f}{\partial v}\right)_{u}\left(\frac{\partial v}{\partial x}\right)_{y}+\left(\frac{\partial f}{\partial u}\right)_{v}\left(\frac{\partial u}{\partial x}\right)_{y}\right]dx+\left[\left(\frac{\partial f}{\partial u}\right)_{v}\left(\frac{\partial u}{\partial y}\right)_{x}+\left(\frac{\partial f}{\partial v}\right)_{u}\left(\frac{\partial v}{\partial y}\right)_{x}\right]dy$
As f is also a function of x and y we can also write:
$df=\left(\frac{\partial f}{\partial x}\right)_{y}dx+\left(\frac{\partial f}{\partial y}\right)_{x}dy$

From the last two equations, we can solve:
$\left(\frac{\partial f}{\partial x}\right)_{y}=\left(\frac{\partial f}{\partial v}\right)_{u}\left(\frac{\partial v}{\partial x}\right)_{y}+\left(\frac{\partial f}{\partial u}\right)_{v}\left(\frac{\partial u}{\partial x}\right)_{y}$
$\left(\frac{\partial f}{\partial y}\right)_{x}=\left(\frac{\partial f}{\partial u}\right)_{v}\left(\frac{\partial u}{\partial y}\right)_{x}+\left(\frac{\partial f}{\partial v}\right)_{u}\left(\frac{\partial v}{\partial y}\right)_{x}$

You can see that this resembles the upper row of your second to last result. If you rework your proof using these rules, it should find the proof you need.

Good luck!