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I have
$$\begin{aligned}x_{1}&=r\sin(\theta_{1}),\\ x_{2}&=r\cos(\theta_{1})\sin(\theta_{2})\\ x_{3}&=r\cos(\theta_{1})\cos(\theta_{2}). \end{aligned} $$
I know how to compute the Jacobian $$\frac{\partial(x_{1},x_{2},x_3)}{\partial(\theta_{1},\theta_{2},r)}$$ directly.

The thing is, there is a way to get this Jacobian that involves a ratio of an upper triangular determinant and a lower triangular determinant. I just cannot figure out how to get this. I'm guessing it's some chain rule thing. Any help will be appreciated. Thanks.

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    You can use $\TeX$ on this site by enclosing formulas in dollar signs; single dollar signs for inline formulas and double dollar signs for displayed equations. You can see the source code for any math formatting you see on this site by right-clicking on it and selecting "Show Math As:TeX Commands". [Here](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference)'s a basic tutorial and quick reference. There's an "edit" link under the question.2012-09-28
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    It is the determinant of a 3 by 3 matrix of partial derivatives of variables $(x, y, z)$ with respect to the other variables $(r, \theta_1, \theta_2)$. I get $r^2\sin(\theta_1)\cos(\theta_2)^2$.2012-09-28
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    That's what I got too - the thing is, the actual problem is for $x_{1}, \cdots, x_{n}$ - that's why that trick I was talking about would be very useful2012-09-28

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