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Partial Derivatives with Inverse Trig Functions

  1. Feb 11, 2008 #1
    [SOLVED] Partial Derivatives with Inverse Trig Functions

    1. The problem statement, all variables and given/known data
    Show that u(x,y) and v(x,y) satisfy the Cauchy-Riemann equations...

    [tex]\frac{\partial u}{\partial x}[/tex] = [tex]\frac{\partial v}{\partial y}[/tex]

    given that u = ln(x[tex]^{2}[/tex] + y[tex]^{2}[/tex])
    and that v = 2tan[tex]^{-1}[/tex] (y/x)

    2. Relevant equations
    single variable differentials of equations should be in the form of..

    [tex]\frac{d}{dx}[/tex] ln (u) = [tex]\frac{1}{u}[/tex] [tex]\frac{du}{dx}[/tex]

    [tex]\frac{d}{dx}[/tex] tan[tex]^{-1}[/tex] u = [tex]\frac{1}{1+u^{2}}[/tex] [tex]\frac{du}{dx}[/tex]
    3. The attempt at a solution
    I would think that the partial derivatives of the above equations would lead from the single differentials, but instead of [tex]\frac{du}{dx}[/tex] there would be [tex]\frac{\partial u}{\partial x}[/tex] or [tex]\frac{\partial u}{\partial y}[/tex] depending on which the question called for. With this reasoning, I thought the left side of the Cauchy-Riemann equation should be..

    [tex]\frac{2x}{x^{2} + y ^{2}}[/tex]

    and the right side would be...

    [tex]\frac{2}{1 + (y/x)^{2}}[/tex] [tex]\frac{1}{x}[/tex]

    However, these don't appear to be equal... unless I can't see the simple algebra. Any help would be greatly appreciated..
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 11, 2008 #2

    NateTG

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    [tex]\frac{2}{1+\left(\frac{y}{x}\right)^2} \times \frac{1}{x}[/tex]
    Distribute through and multply by 1 ([itex]\frac{x}{x][/itex])...
    [tex]\frac{2}{x+\frac{y^2}{x}} \times \frac{x}{x}[/tex]
    [tex]\frac{2x}{x^2+y^2}[/tex]
     
  4. Feb 11, 2008 #3
    in the denominator of your responce you have the term y^{2}/x. Shouldn't the term be y^{2}/x^{2}?
     
  5. Feb 11, 2008 #4

    NateTG

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    [tex]\frac{1}{1+\left(\frac{y}{x}\right)^2} \times \frac{1}{x}[/tex]
    [tex]\frac{1}{\left(1+\left(\frac{y}{x}\right)^2\right) x}[/tex]
    [tex]\frac{1}{x + x \times \frac{y^2}{x^2}}[/tex]
    [tex]\frac{1}{x + \frac{y^2}{x}}[/tex]
     
  6. Feb 11, 2008 #5
    wow... i feel pathetic... thanks again.
     
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