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## Homework Statement

Find [tex] \frac{\partial z}{\partial u} [/tex] and [tex] \frac{\partial z}{\partial v} [/tex] using the chain rule.

[tex] z = \arctan(\frac{x}{y}) , x=u^2+v^2 , y=u^2-v^2 [/tex]

## Homework Equations

## The Attempt at a Solution

[tex] \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)} [/tex]

[tex] \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)} [/tex]

why is my answer wrong?

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