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Homework Help: Double derivative proof

  1. Sep 29, 2010 #1
    1. The problem statement, all variables and given/known data
    [tex]\sqrt{x}+\sqrt{y}= 4[/tex] prove that [tex]y''=\frac{2}{x\sqrt{x}}[/tex]



    3. The attempt at a solution
    [tex]\frac{dy}{dx}=-\frac{\sqrt{y}}{\sqrt{x}}[/tex]

    I tried solving for the second derivative and got

    [tex]\frac{d^2y}{dx^2}=-\frac{-x\frac{dy}{dx}-y}{2\sqrt{\frac{y}{x}}x^2}[/tex]
    [
    Which is wrong if I plug in values.
    Anyone sees my error?
     
  2. jcsd
  3. Sep 29, 2010 #2

    danago

    User Avatar
    Gold Member

    Your first derivative looks ok to me, but im not too sure where your second derivative came from. Did you use the quotient rule?

    [tex]
    \frac{d}{{dx}}(\frac{u}{v}) = \frac{{u'v - v'u}}{{v^2 }}
    [/tex]
     
  4. Sep 1, 2012 #3
    How did you get [itex]\frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}}[/itex]?
     
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