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Partial derivative problem

  1. Aug 7, 2012 #1
    Hi!
    Here is my function:

    2ag2fia.jpg

    My task is to find:

    206irmv.jpg

    I think I know how to find ∂u/∂x, but I have no idea how to find ∂/∂z(∂u/∂x). Here is how I found ∂u/∂x:

    http://oi48.tinypic.com/prsly.jpg

    Does someone know how to find ∂/∂z(∂u/∂x)?
    I appreciate any help :)
     
    Last edited by a moderator: Aug 8, 2012
  2. jcsd
  3. Aug 7, 2012 #2
    Before the second partial derivative, you should fix the error in your calculation of ∂u/∂x, specifically ∂([itex]\frac{xy}{z}[/itex])/∂x.
     
  4. Aug 7, 2012 #3
    What's wrong with ∂(xy/z)/∂x? I checked it and it seems correct to me...
     
  5. Aug 8, 2012 #4
    It's very important so all suggestions are welcome :)
     
  6. Aug 8, 2012 #5
    Never mind. I hadn't scrolled all the way down, it is correct.

    I believe you are having trouble calculating [itex]\frac{∂}{∂z}[/itex]([itex]∂\rho/∂s[/itex]) and [itex]\frac{∂}{∂z}[/itex]([itex]∂\rho/∂t[/itex]) (Let me know if this is not the case).
    To simplify this, get rid of s and t by writing [itex]∂\rho/∂s[/itex] and [itex]∂\rho/∂t[/itex] as partial derivatives of [itex]\rho[/itex] w.r.t. x, y and z, using the chain rule. Since you know how s and t depend on x, y and z, this can be done.

    Once you have done this, calculating [itex]\frac{∂}{∂z}[/itex]([itex]∂\rho/∂s[/itex]) and [itex]\frac{∂}{∂z}[/itex]([itex]∂\rho/∂t[/itex]) would be straightforward.
     
  7. Aug 8, 2012 #6
    I got it finally :smile: Thaks a lot!
     
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