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Homework Help: Partial derivatives

  1. Sep 19, 2005 #1
    On question

    If I have the plane z = y + f(x,y) where f(x,y) = sin(x) * y

    Is it possible to find the complete partial derivatives for z ?

    /Bob
     
  2. jcsd
  3. Sep 19, 2005 #2

    TD

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    "Complete partial derivatives"?

    Perhaps you mean the total derivative?
     
  4. Sep 19, 2005 #3
    Help

    Yes,

    I made a typoo in my last post:

    What I meant to write was

    I'm presented with [tex] z = y + f(x^2 - y^2) [/tex]

    I'm told this can be written as [tex] y \frac{\partial z}{\partial x} + x \frac{\partial z}{\partial y} = x[/tex]

    Any idears on how I do that ?

    /Bob

    p.s. I get [tex]y \frac{\partial x}{\partial x} + \frac{\partial z}{\partial y} + 2x - y^2 \frac{\partial z}{\partial x} + x^2 -2y \frac{\partial z}{\partial y} = 0[/tex]

    But how do I go from this result to the expected result ?
     
    Last edited: Sep 20, 2005
  5. Sep 20, 2005 #4

    Fermat

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    using the z-function as now given,

    get [tex]\frac{\partial z}{\partial x}[/tex] and [tex]\frac{\partial z}{\partial y}[/tex]
    Create the lhs using these values of [tex]\frac{\partial z}{\partial x}[/tex] and [tex]\frac{\partial z}{\partial y}[/tex] and you will end up with the rhs. viz [itex]x[/itex]

    Edit: it's fairly simple - not complicated.
     
  6. Sep 20, 2005 #5
    Okay

    I then get

    [tex]\frac{\partial z}{\partial x} = 2x [/tex]
    [tex]\frac{\partial z}{\partial y} = - 2y[/tex]

    What do I then do next ?

    /Bob
     
    Last edited: Sep 20, 2005
  7. Sep 20, 2005 #6

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    Those partial derivatives are quite wrong, I'm afraid.

    e.g. the partial derivative of y², wrt x, is zero!!

    z = y + f(x²- y²)

    dz/dx = 0 + (f'')*(2x) where f' is the derivative of f wrt its argument.

    and dz/dx is the partial derivative.
     
  8. Sep 20, 2005 #7

    Fermat

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    Here's how to do those partial derivatives.

    [tex]\mbox{If }z = f(\phi),\mbox{ where } \phi = x^2 + y^2[/tex]

    then

    [tex]\frac{\partial z}{\partial x} = \frac{\partial f}{\partial \phi}.\frac{\partial \phi}{\partial x}[/tex]

    where

    [tex]\frac{\partial f}{\partial \phi} \mbox{ is } f'[/tex]

    and

    [tex]\frac{\partial \phi}{\partial x} = 2x + 0[/tex]

    So,

    [tex]\frac{\partial z}{\partial x}= f'.(2x)[/tex]
     
  9. Sep 20, 2005 #8

    Fermat

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    Not quite right yet.
    Have you seen my earlier post?

    Bob, could you post a new message rather than sinmply editing an old post? No one can tell if you have responded or not unless they actually read the old message. If you post a new reply, it will show up as such in the forum index page.
    Thanks
     
  10. Sep 20, 2005 #9
    Yes sorry,

    Please excuse me if mix up standard differentation and partial differentation

    If [tex]z = y + f(x^2 - y^2)[/tex]

    Can be written as:

    [tex]y \frac{\partial z}{\partial x} + x \frac{\partial z}{\partial y} = x[/tex]

    Then to use the notation required in the task [tex]\frac{\partial z}{\partial x}= 2x \frac{\partial f}{\partial x}[/tex]

    [tex]\frac{\partial z}{\partial y}= 1- 2y \frac{\partial f}{\partial y}[/tex]

    If this is correct then how do I go from these two deriatives to the required result ?

    Sincerely

    /Bob
     
    Last edited: Sep 20, 2005
  11. Sep 20, 2005 #10

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    I'm afraid these partial derivatives are still wrong.

    Have you read my earlier post, post #7, showing how to do the partial derivatives.

    There should be a [itex]f'[/itex] in the expression for [tex]\partial z/\partial x[/tex]
     
  12. Sep 20, 2005 #11
    Is it then ??

    [tex]\frac{\partial z}{\partial x}= f'. 2x[/tex]

    [tex]\frac{\partial z}{\partial y}= f'. 1- 2y[/tex]

    /Bob
     
  13. Sep 20, 2005 #12

    Fermat

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    Almost there,

    [tex]\mbox{The }\frac{\partial z}{\partial y} \mbox{ should be }\frac{\partial z}{\partial y}= 1- f'.2y[/tex]

    [tex]z = y + f(\phi) \mbox{ where } \phi = x^2 - y^2[/tex]

    [tex]\frac{\partial z}{\partial y} = \frac{\partial y}{\partial y} + \frac{\partial f}{\partial \phi}.\frac{\partial \phi}{\partial y}[/tex]
    [tex]\frac{\partial z}{\partial y} = 1 + f'.(-2y)[/tex]
    [tex]\frac{\partial z}{\partial y} = 1 - f'.(2y)[/tex]
     
  14. Sep 20, 2005 #13
    Okay

    Thank You very much :-)

    My final question is than which approach do I have use to obtain that

    [tex]z = y + f(x^2 - y^2)[/tex]

    can be written as

    [tex] y \frac{\partial z}{\partial x} + x \frac{\partial z}{\partial y} = x [/tex]

    Do I add the two partial deriatives together ?

    Best Regards

    Bob

     
  15. Sep 20, 2005 #14

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    [tex]\frac{\partial z}{\partial x} = 2x.f'[/tex]

    and

    [tex]\frac{\partial z}{\partial y} = 1 - 2y.f'[/tex]

    So,

    [tex]x.\frac{\partial z}{\partial y} + y.\frac{\partial z}{\partial y} = x.(1 - 2y.f') + y.2x.f' = ??[/tex]
     
  16. Sep 20, 2005 #15
    Thank You again,

    I can see that now that, those two equal each other, but how does it equal x ?

    /Bob

     
  17. Sep 20, 2005 #16

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    Bob, can you not see that,

    [tex]x.(1 - 2y.f') + y.2x.f' = x - 2xy.f' + 2xy.f' = x\ ?[/tex]
     
  18. Sep 20, 2005 #17
    Now I can ;-)

    Thank You ;-)


    /BOb

    Thank You

     
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