Find Partial Derivatives of z with Respect to u and v

[-EquinoX-]

Homework Statement

Find $$\frac{\partial z}{\partial u}$$ and $$\frac{\partial z}{\partial v}$$ using the chain rule.

$$z = \arctan(\frac{x}{y}) , x=u^2+v^2 , y=u^2-v^2$$

Homework Equations

The Attempt at a Solution

$$\frac{\partial z}{\partial u} = \frac{4uv^2}{v^4 - 2u^2v^2 + u^4} * \frac{1}{1+((u^2+v^2)/(u^2-v^2))^2)}$$

$$\frac{\partial z}{\partial v} = \frac{4u^2v}{v^4 - 2u^2v^2 + u^4} * \frac{1}{(1+((u^2+v^2)/(u^2-v^2))^2)}$$

why is my answer wrong?

 

[quantumdude]

why is my answer wrong?

Are we supposed to guess at what steps are written on your paper?

 

[Dick]

For one thing d(arctan(x))/dx=1/(1+x^2). I think you missed the square.

 

[-EquinoX-]

oh you're right, I missed the squares, but it is still wrong

 

[Dick]

Tom Mattson is right. It would be a lot easier to say what is wrong if you would show what you've done. Just showing a wrong answer and saying 'what did I do wrong?' is more of a puzzle than a question. I don't think you've done much wrong. I get a different sign for dz/du and you could certainly simplify them more.

 

[-EquinoX-]

well I think the problem here is just from taking the derivative of

$$\frac{u^2+v^2}{u^2-v^2}$$ with respect to u, right. I realize that there should be a - sign in front of it.. am I right?

 

[Dick]

Yes. But like I said you can also simplify those expressions a lot. I don't know if the HW checker requires this or not.

 

[-EquinoX-]

the best simplification I can think of is:

$$\frac{-4uv}{(u^2-v^2)^2+(u^2+v^2)^2}$$

 

[Dick]

Do you mean -4uv/((u^2-v^2)^2+(u^2+v^2)^2)?? I can write that in less space.

 

[-EquinoX-]

Yes that's what I meant