Chain rule

  • Thread starter -EquinoX-
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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?
 
Last edited:

Answers and Replies

  • #2
Tom Mattson
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why is my answer wrong?

Are we supposed to guess at what steps are written on your paper? :confused:
 
  • #3
Dick
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For one thing d(arctan(x))/dx=1/(1+x^2). I think you missed the square.
 
  • #4
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oh you're right, I missed the squares, but it is still wrong
 
  • #5
Dick
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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
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well I think the problem here is just from taking the derivative of

[tex] \frac{u^2+v^2}{u^2-v^2} [/tex] with respect to u, right. I realize that there should be a - sign in front of it.. am I right?
 
  • #7
Dick
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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
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the best simplification I can think of is:

[tex] \frac{-4uv}{(u^2-v^2)^2+(u^2+v^2)^2} [/tex]
 
Last edited:
  • #9
Dick
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Do you mean -4uv/((u^2-v^2)^2+(u^2+v^2)^2)?? I can write that in less space.
 
  • #10
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Yes that's what I meant
 

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