# 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).