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Question about partial derivatives with three unknown

  1. Apr 9, 2013 #1
    1. The problem statement, all variables and given/known data

    If z=f(x,y) with u= x^2 -y^2 and v=xy , find the expression for (∂x/∂u).
    the (∂x/∂u) will be used to calsulate ∂z/∂u.
    my question is how to find (∂x/∂u).
    I don't know what to keep constant. Maybe the question has some problem.
    The answer is (∂x/∂u)=(x/2)/(x^2+y^2).
    Can you please guide me .


    2. Relevant equations



    3. The attempt at a solution
    I use u= x^2 -y^2, keep y constant , then get du/dx=2x , then 1/(du/dx)
    or keep y constant, then x=√(u+y^2), get dx/du= 1/2x?

    Or I should not keep anything constant? or the question is wrong?

    really thank
     
  2. jcsd
  3. Apr 9, 2013 #2

    Simon Bridge

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    ##u=x^2-y^2 \Rightarrow x^2 = u+y^2 \\ \Rightarrow \frac{\partial}{\partial u}(x^2 = u+y^2)=\cdots## ... that help?
     
  4. Apr 9, 2013 #3
    keep y constant?? if so the answer is 1/2x also.
    if not keeping y constant ......
    I really dont know how to get this (∂x/∂u)=(x/2)/(x^2+y^2).
     
  5. Apr 9, 2013 #4

    Simon Bridge

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    $$\frac{\partial}{\partial u}(x^2 = u+y^2)= \frac{\partial}{\partial u}x^2= 1 + \frac{\partial}{\partial u}y^2$$ ... should you keep y constant?
    I think that's what you need to decide - is it ##x(y,u)=u+y(x,u)##?
    Definition of the partial derivative?
     
  6. Apr 9, 2013 #5
    Quote from wolfram math world:
    Partial derivatives are defined as derivatives of a function of multiple variables when all but the variable of interest are held fixed during the differentiation.
    ∂y/∂u , y=f(u,x), since u varies here, so y will vary. Can you please guide how should I calculate ∂y/∂u?
    Thank you.
     
  7. Apr 9, 2013 #6

    Simon Bridge

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    $$u=x^2-y^2$$ ... and you want to show that $$\frac{\partial x}{\partial u}=\frac{x}{2(x^2+y^2)}$$... is that correct?
     
  8. Apr 9, 2013 #7
    Yup, please
     
  9. Apr 9, 2013 #8

    Simon Bridge

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    Normally when you have y=f(x) then dx/dy = 1/(dy/dx)
    Does this work for partials?

    $$\frac{\partial}{\partial x}u(x,y) \Rightarrow \left . \frac{\partial u}{\partial x} \right |_y$$

    If you consider x to be a function of u and y, then the partial - holding y constant - cannot come out the way you want.

    There is a lot more information about the problem that is not provided.
     
  10. Apr 9, 2013 #9
    This only applies to partial , is not? I learned that from thermo.

    If the question says find (∂x/∂u) keeping y constant , then we can use du/dx = 1/(dx/du) to get the answer without care about y=f(u,x) because u depends on x and y , but x and y are independent. the question no need to give us a equation of y in term of x to calculate dy/dx.
    Is that correct?
     
  11. Apr 9, 2013 #10
    Here is the therma's proof
     

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  12. Apr 9, 2013 #11

    Simon Bridge

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    That's the idea ... $$\left . \frac{\partial x}{\partial u}\right |_y = \left ( \left . \frac{\partial u}{\partial x}\right |_y \right )^{-1}$$ ... works because the same variable is being held constant.

    However: $$u=x^2-y^2 \Rightarrow \frac{\partial u}{\partial x} = 2x$$... like you already noticed but this does not give you what you need.

    It follows that there is something else going on.
     
  13. Apr 9, 2013 #12
    Is the question not complete, it should be (∂x/∂u) keeping v constant.
    Thank for wasting so much time on me. Really sorry.
     
  14. Apr 9, 2013 #13

    Simon Bridge

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    You said, post #1, you didn't know what to keep constant.
    ... nothing there about keeping v constant - I wondered why it was relevant.

    Ideally you want to express x as a function of u and v then.
    Maybe express u as a function of x and v and evaluate du/dx|v and invert it.
     
  15. Apr 10, 2013 #14
    Yup ,can get answer from that. Thank you.
     
  16. Apr 10, 2013 #15

    Simon Bridge

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    Well done - sometimes you don't see what's there until you have to talk about it to someone else :)
     
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