# Chain rule

1. Feb 21, 2009

### -EquinoX-

1. The problem statement, all variables and given/known data
Find $$\frac{\partial z}{\partial u}$$ and $$\frac{\partial z}{\partial v}$$ using the chain rule.

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

2. Relevant equations

3. 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?

Last edited: Feb 21, 2009
2. Feb 21, 2009

### Tom Mattson

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

3. Feb 21, 2009

### Dick

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

4. Feb 21, 2009

### -EquinoX-

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

5. Feb 21, 2009

### 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.

6. Feb 21, 2009

### -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?

7. Feb 21, 2009

### 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.

8. Feb 21, 2009

### -EquinoX-

the best simplification I can think of is:

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

Last edited: Feb 21, 2009
9. Feb 21, 2009

### Dick

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

10. Feb 21, 2009

### -EquinoX-

Yes that's what I meant