1. ## Jacobian determinant question

The equations,

$\displaystyle x^{3} + xy - uy = v + 4,$ $\displaystyle y^{2} - xy^{3} + uv = 1,$

define x and y as functions of u and v. Find, $\displaystyle \frac{\partial{x}}{\partial{v}}$

See figure attached for my attempt. It get's a little messy with the bottom determinant, so I'm not entirely convinced I've done this correctly.

Did I do anything wrong?

2. Originally Posted by jegues
The equations,

$\displaystyle x^{3} + xy - uy = v + 4,\;\;y^{2} - xy^{3} + uv = 1,$

define x and y as functions of u and v. Find $\displaystyle \frac{\partial{x}}{\partial{v}}$

See figure attached for my attempt. It get's a little messy with the bottom determinant, so I'm not entirely convinced I've done this correctly.

Did I do anything wrong?
Doing it by a slightly different method, I got almost the same answer. In fact, if you use implicit differentiation to differentiate each of the given equations partially with respect to v then you get

$\displaystyle 3x^2\frac{\partial{x}}{\partial{v}} + y\frac{\partial{x}}{\partial{v}} + x\frac{\partial{y}}{\partial{v}} -u\frac{\partial{y}}{\partial{v}} = 1,$
$\displaystyle 2y\frac{\partial{y}}{\partial{v}} -y^3\frac{\partial{x}}{\partial{v}} - 3xy^2\frac{\partial{y}}{\partial{v}} + u = 0.$

Write these equations as

$\displaystyle (3x^2+y)\frac{\partial{x}}{\partial{v}} + (x-u)\frac{\partial{y}}{\partial{v}} = 1,$
$\displaystyle y^3\frac{\partial{x}}{\partial{v}} + (3xy^2-2y)\frac{\partial{y}}{\partial{v}} = u.$

Multiply the first equation by $\displaystyle 3xy^2-2y$ and the second by $\displaystyle x-u$ and subtract, to get

$\displaystyle \bigl((3x^2+y)(3xy^2-2y) - y^3(x-u)\bigr)\frac{\partial{x}}{\partial{v}} = 3xy^2-2y - (x-u)u.$

That is the same as your answer except that one of us has missed a minus sign somewhere, so that my denominator is the negative of yours.