Problem finding a partial derivative

[chexmix]

Homework Statement

I am working on a homework problem involving partial derivatives. I've been checking my answers against what Wolfram Alpha spits out just for extra assurance. For the following problem

Find all the second partial derivatives: v = $\frac{xy}{(x-y)}$.

When I get to the point where I am deriving v$_{xy}$, Wolfram Alpha gives me

$\frac{2xy}{(x-y)^3}$

but I get something different. If someone can tell me where I am going wrong here, I would be most grateful: I've stared at this for some time now. I suspect it is something very trivial. I am using the quotient rule.

2. The attempt at a solution

$\frac{\partial}{\partial y}$ $\frac{y^2}{(x-y)^2}$

$\frac{(x-y)^2 * 2y - y^2 * (2 (x-y))}{(x-y)^4}$

$\frac{(x-y)((x-y)*2y - 2y^2)}{(x-y)^4}$

$\frac{2xy-4y^2}{(x-y)^3}$

Thanks,

Glenn

 

[Bassalisk]

You forgot the minus.
$\frac{\partial}{\partial y}$ $\frac{-y^2}{(x-y)^2}$

And in your quotient rule, you forgot the derivative on the "inner" function, which is -1.

$\frac{\partial(x-y)^{2}}{\partial y}=2(x-y)(-1)$

 

[chexmix]

Ahh, I was even further off than I thought. Thanks so much. It makes sense now.