Partial derivatives

1. Apr 3, 2009

Pietair

1. The problem statement, all variables and given/known data
If z = 1 / (x^2+y^2-1)
show that x(dz/dx)+y(dz/dy)=2z(1+z)

2. The attempt at a solution
z = (x^2+y^2-1)^-1
dz/dx = -2x(x^2+y^2-1)^-2 = -2x * z^2
dz/dy = -2y(x^2+y^2-1)^-2 = -2y * z^2

(-2x^2 * z^2) - (2y^2 * z^2) = 2z(1+z)

I can express x and y in something like z and x/y:
x = z^-1 - (z^-1*y^2)
y = z^-1 - (z^-1*x^2)

Though substituting this values in the obtained equation doesn't get me near the answer...

2. Apr 3, 2009

HallsofIvy

Staff Emeritus
Yes, exactly right. Why would you want to do the following?

3. Apr 3, 2009

Pietair

Because I have no idea how to get rid of the x and y...

4. Apr 3, 2009

n!kofeyn

There's a negative sign missing in the formula you need to show. It should be
$$x \frac{dz}{dx} + y \frac{dz}{dy} = -2z(1+z)$$

5. Apr 4, 2009

n!kofeyn

It sounds like you're trying to get x and y in terms of z once you have found the partial derivatives. Leave the partials in terms of x and y, and then expand out the -2z(1+z) in terms of what z equals. This will help you verify the formula (please see my previous post for the correction to the problem statement).