How do I Find the Jacobi Matrix and Its Determinant for a Given Transformation?

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Homework Statement



The transformation f is defined by: R^2 --> R^2 and is defined by:
f(x,y) = (y^5, x^3)

Find the jacobi matrix and its determinant

Homework Equations



f(x,y) = (y^5, x^3)

The Attempt at a Solution



I would start by differentiating y^5 with respect to x and then y, then differentiate x^3 with respect to x and then y.

I end up with:

Df = (0... 5y^4 )
...(3x^2... 0 )

And from here I don't know what to do. Usually I would be told, that Df = (2,1) for example, and then I would place them instead of y and x, but here I am not given any other information than what I have written above. How do I find the determinant?
 
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How about just doing what you were told to do? Yes, if you were asked to find the Jacobian matrix and its determinant at, say, (1, 2) you would replace x and y with those. But you aren't asked to do that so just leave them as "x" and "y".
 
HallsofIvy said:
How about just doing what you were told to do? Yes, if you were asked to find the Jacobian matrix and its determinant at, say, (1, 2) you would replace x and y with those. But you aren't asked to do that so just leave them as "x" and "y".

Okay, thank you.

But I have another question, what is the difference between the Jacobi Matrix and the Jacobi determinant?
 
I am also asked to:

let D be the set of points (x,y) in R^2 doe which 0 ≤ x ≤ 1 and

0 ≤ y ≤ 1. Find a function g: R^2 --> R for which

1010 h(x,y)dxdy =
1010 h(y5, x3*g(x,y)dxdy

is true for all functions h: D --> R integrable over D
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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