In this book I'm studying I've come across a problem where the author solves a partial differentiation problem using determinants. I'm somewhat familiar with them, but I don't see how they derive the answer from the two they started with:
\begin{cases} v+log\left|u\right|=xy \\ u+log\left|v\right|=x-y \end{cases} \begin{cases} \frac{1}{u}\frac{\partial{u}}{\partial{x}}+\frac{\partial{v}}{\partial{x}}=y \\ \frac{1}{v}\frac{\partial{v}}{\partial{x}}+\frac{\partial{u}}{\partial{x}}=1 \end{cases}
So in retrospect they started with the first bracket above with two equations in it and the four variables v, u, x and y, and then used implicit partial differentiation to arrive at the second bracket.
Are dependent and independent variables treated differently when taking the derivative?
Lastly, they moved into lesser-known territory for me and solved the second bracket for $\frac{\partial{u}}{\partial{x}}$:
$$ \frac{\partial{u}}{\partial{x}}=\frac{ \begin{vmatrix} yu & u \\ v & 1 \\ \end{vmatrix}} {\begin{vmatrix} 1 & u \\ v & 1 \\ \end{vmatrix}}= \frac{{u}\left(y-v\right)}{1-uv}$$
How is it that they go from the bracket to that determinant?
Thank you, Brandon
Some remarks:
Take into account that $x$ and $y$ seem to be the indepedent variables of your problem. Then, it turns out that: $$ \frac{\partial (xy)}{\partial x} = \frac{\partial (x)}{\partial x} y + x\frac{\partial (y)}{\partial x} = y, \quad \frac{\partial (x-y)}{\partial x} = \frac{\partial (x)}{\partial x} - \frac{\partial (y)}{\partial x} = 1,$$ since $\partial_x y = 0$.
In order to come up with the expression for $u_x$ and $v_x$, the Cramer's rule has been used. Note that you can write the second bracket as follows: $$\left(\begin{array}{cc} 1 & u \\ v & 1 \\ \end{array}\right) \mathbf{U} = \left(\begin{array}{c} y u \\ v \\ \end{array}\right), $$ where $\mathbf{U}=(\partial_x u , \partial_x v)$ and we have multiplied the first row by $u$ and the second by $v$.
I hope this helps.
Cheers!