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Homework Help: Superior implicit differentiation, prove answer.

  1. Sep 28, 2010 #1
    1. The problem statement, all variables and given/known data
    [tex]2XY = Y^2 prove that y''2 = \frac{y^{2}-2xy}{(y-x)^3}[/tex]
    EDIT: Sorry, don't know how to insert a space in Latex.
    2. Relevant equations



    3. The attempt at a solution
    [tex] 2y+2x \frac{dy}{dx} = 2y \frac{dy}{dx} [/tex]
    [tex] \frac{dy}{dx}(2y-2x) = 2y[/tex]
    [tex]\frac{dy}{dx}= \frac{y}{y-x} [/tex]
    [tex]\frac{d^2y}{dx^2}=\frac{\frac{dy}{dx}(y-x)-(\frac{dy}{dx}-1)y}{(y-x)^2} [/tex]
    [tex]\frac{d^2y}{dx^2}= \frac{y^2}{(y-x)^3}[/tex]


    Is my work correct? If yes, then the question itself is wrong.
    In the original equation (2XY = Y2) does it matters that the letters are caps?
     
  2. jcsd
  3. Sep 28, 2010 #2

    Dick

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    Your work is correct up to the last line. But the given solution is the correct one. Looks like something went wrong when you substituted your expression for dy/dx in and simplified.
     
  4. Sep 28, 2010 #3
    Plugging in gives me:
    ((y/y-x)(y-x)-((y/y-x)-1)y)/((y-x)2)
    Simplifiying:
    (y-(y2/(y-x))-y))/((y-x)2)
    (-y2/(y-x))/((y-x)2)

    Different than my first try (now it's negative), but still I don't know how to the the -2xy in there.
     
  5. Sep 28, 2010 #4

    Dick

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    In y-(y^2/(y-x)-y) the y's don't cancel. It's minus of a minus.
     
  6. Sep 28, 2010 #5
    Got it, my mistake was that I moved the (y-x) that was in the numerator as a denumerator before adding fractions. And the wrong minus of a minus that you pointed.
    Thanks.
     
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