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Partial Derivatives of natural log

  1. Nov 12, 2008 #1
    Hey all. I'm having some problems with the partial derivatives of e. I understand the basics such as exy2. where I'm getting confused is with the following

    dz/dx=e(x+y)

    and

    dz/dx=1/ex+ey

    Can any one help me out with understanding these??
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 12, 2008 #2

    HallsofIvy

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    I don't understand what you are asking. You ask about finding the derivative but then give the derivative. Are asking for the partial derivatives of ex+y and 1/(ex+ ey or are you saying these are the derivatives and you want to find z?

     
  4. Nov 14, 2008 #3
    I'm sorry for being confusing. Yes I'm thrying to find the partial derivative of each item. I have no problems with partial derivatives that dont contain "e" but these I am having problems grasping.
     
  5. Nov 14, 2008 #4

    HallsofIvy

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    Do you know what the derivative of ex is? The ordinary derivative, not partial derivative. It's the world's easiest derivative!

    Do you know what the derivative of Cex is? ex+y is exey.

    1/ex+ ex= e-x+ ex.

    Those are both easier than [itex]e^{xy^2}[/itex]!
     
  6. Nov 16, 2008 #5
    I guess I'm not asking this very well. I'm trying to solve the problem in parts but that's not helping me understand this any better. The full problem is...

    z=(ex+y)/(ex+ey)

    And I need to find the partial derivatives in respect to "x" and "y"
    I am coming up with an answer that fully cancels out and winds up being zero when added back together.

    Here's my work...

    dz/dx=(ex+ey)(exey)-(exey)(ex+ey)/(ex+ey)2
    dz/dy= the same as above.

    I'm sure I'm missing something, but can't find where. Can you help please?
     
  7. Nov 17, 2008 #6

    HallsofIvy

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    It is always a good idea to state the entire problem!

    The derivative of exey with respect to x is just exey again and I think you have that in your answer. But the derivative of ex+ ey is just ex because ey is a constant with respect to x and its derivative is 0.

    dz/dx=(ex+ey)(exey)
    is correct
    -(exey)(ex+ey)
    this is wrong. You should have -(exey)(ex)
    (the denominator was correct so I didn't mention it.)
     
  8. Nov 17, 2008 #7
    Thank you, now that you pointed out where i was wrong, it makes much more sense!:biggrin:
     
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