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Partial Differentiation

  1. Dec 2, 2009 #1
    I was working through this problem,

    I understand the method, but got stuck trying to differentiate the arc tan terms...

    In the table of standard derivatives,

    [tex]\frac{d}{dx} \arctan{x} = \frac{1}{1+x^2}[/tex]

    and for PDE's, you treat y as constant when differentiating w.r.t. x

    ---

    Any help greatly appreciated!
    :)
     

    Attached Files:

  2. jcsd
  3. Dec 2, 2009 #2

    Mark44

    Staff: Mentor

    When you are doing partial differentiation, you treat each variable as independent from the others. So when you find the partial of your function w.r.t. x, say, you treat y as a constant.
     
  4. Dec 2, 2009 #3
    I get that - if you could have a look at my attachment, that has the question in it :)
     
  5. Dec 2, 2009 #4

    Mark44

    Staff: Mentor

    Suppose you had u = u(x, y) = x2tan-1(y/x), just to make it easier for me to type this in.

    [tex]\frac{\partial u}{\partial x}~=~2x~tan^{-1}(y/x)~+~x^2~\frac{\partial}{\partial x}\left(tan^{-1}(y/x)\right)[/tex]
    [tex]=~2x~tan^{-1}(y/x)~+~x^2~\frac{1}{1 + (y/x)^2}~\frac{\partial}{\partial x}\left(\frac{y}{x}\right)[/tex]
    For that last partial, you have what amounts to y*d/dx(1/x^2) = -y/x.

    Clear?
     
  6. Dec 6, 2009 #5
    Yeah, I just missed a bit of chain-rule didn't I?
    Thanks for the help :)
     
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