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Partial Differentiation

  1. Oct 8, 2005 #1
    Hi, can anyone help me with the following differentiation question?

    Find first partial derivatives w.r.t to x and y for:

    tan (x/y)

    Can anyone offer any help on how to approach this question? I know when you differentiate (x/y) using the product rule, that you have to differentiate y implicitly...but how does that work when using partial derivatives? :confused:

    Thanks for taking time to read :smile:
     
  2. jcsd
  3. Oct 8, 2005 #2

    mezarashi

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    In partial derivatives, you treat all variables as constants (you know, the a, b, c)... except for the one you are differentiating with respect to.

    For example:

    f(x,y,z) = 2xyz + 3x + 4y + 5z
    df/dz = 2xy + 5
    df/dx = 2yz + 3
    df/dy = 2xz + 4
     
  4. Oct 9, 2005 #3
    Ok, I've given it a go :uhh: but I'm not sure if its right...

    f (x,y) = tan (x/y)

    Let f(x,y) = tan (u) where u = (x/y) = xy^(-1) *** Not sure if I've used the right method for this part

    f1 (w.r.t x) = (1/y) sec^2 (x/y)

    f2 (w.r.t y) = -(x/y^2) sec^2 (x/y)

    Hmm can someone let me know if its right? :uhh:
     
  5. Oct 9, 2005 #4

    arildno

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    Yes, you have gotten the correct answers.
     
  6. Oct 9, 2005 #5
    Thank you! Glad to know Im on the right track :biggrin:

    If I had to use partial differentiation for (1/2) ln (x^2 + y^2), can I just use the same technique (chain rule) and:

    Let (1/2) ln (x^2 + y^2) = 1/2 ln (u) where u = x^2 + y^2

    Then:

    f1 (w.r.t x) = x ln (x^2 + y^2)

    f2 (w.r.t y) = y ln (x^2 + y^2) ?
     
  7. Oct 9, 2005 #6

    arildno

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    Certainly you can do that.
     
  8. Oct 9, 2005 #7
    ok thanks! There is just one last part of da question regarding the logs which I don't understand.

    Show that x f(1) + y (f2) = 1? :confused:

    I've started with:

    x^2 ln (x^2 + y^2) + y^2 ln (x^2 + y^2)

    so how do I show that it is equal to 1?
     
  9. Oct 9, 2005 #8

    HallsofIvy

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    I don't understand this. Is "x f(1)+ y(f2)= 1" supposed to be
    xfx+ yfy= 1?

    Are you saying that f(x,y)= x2ln(x2+y2)+ y2ln(x2+y2)?

    If those are what you mean, then you can't show that
    xfx+ yfy= 1, it's not true.
     
  10. Oct 9, 2005 #9

    arildno

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    LEARN PROPER NOTATION!!!!!
    :grumpy: :grumpy: :grumpy: :grumpy: :grumpy: :grumpy: :grumpy: :grumpy:
    Now, f(1) as you call it should be [tex]\frac{\partial{f}}{\partial{x}}=\frac{x}{x^{2}+y^{2}}[/tex] and f(2) is really [tex]\frac{\partial{f}}{\partial{y}}=\frac{y}{x^{2}+y^{2}}[/tex]
    when [tex]f(x,y)=\frac{1}{2}ln(x^{2}+y^{2})[/tex]
     
  11. Oct 9, 2005 #10
    Yes it is, I dunno how to do the superscript x you see :blushing:

    Erm I've attached the question here, though its suppose to be 'u' not 'f'

    http://tinypic.com/eg5mhi.jpg
     
  12. Oct 9, 2005 #11

    arildno

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    Just click on the Latex images to see how the code is for the various symbols.
     
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