# Chain rule

## Homework Statement

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

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

## 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?

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## Answers and Replies

Tom Mattson
Staff Emeritus
Gold Member
why is my answer wrong?

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

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

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

Dick
Homework Helper
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.

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
Homework Helper
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.

the best simplification I can think of is:

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

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Dick