Partial Derivatives: Solving x(dz/dx)+y(dz/dy)=2z(1+z)

[Pietair]

Homework Statement

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...

 

[HallsofIvy]

Homework Statement

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)

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

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...

 

[Pietair]

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

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

Thanks for your quick reply.

 

[n!kofeyn]

Homework Statement

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)

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

 

[n!kofeyn]

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

Thanks for your quick reply.

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).