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Derivative Question

  1. May 6, 2007 #1
    1. The problem statement, all variables and given/known data
    For the Relation defined by x^5 + y^5 = 5 show that d2y/dx2.

    2. Relevant equations



    3. The attempt at a solution

    x^5 + y^5 = 5
    5x^4 + 5^4dy/dx = 0
    d2y/dx2 = - 20x^3/20y^3

    ??
     
  2. jcsd
  3. May 6, 2007 #2
    [tex]x^5+y^5=5[/tex]
    [tex]5x^4dx+5y^4dy=0[/tex]
    [tex]5x^4dx=-5y^4dy[/tex]
    [tex]-\frac{5x^4}{5y^4}=\frac{dy}{dx}[/tex]
    [tex]\frac{5y^4(-20x^3)+5x^4(20y^3)}{25y^8}=\frac{d^2y}{dx^2}[/tex]

    [tex]\frac{d^2y}{dx^2}=\frac{100x^4y^3-100x^3y^4}{25y^8}[/tex]
     
    Last edited: May 6, 2007
  4. May 7, 2007 #3
    The answer says d^2y/dx^2 = -20x^3/y^9.

    ??
     
  5. May 8, 2007 #4
    d/dx(x^5) + d/dx(y^5) = d/dx(5)
    5x^4 + 5y^4 * y' = 0
    5x^4 = -5y^4 * y'
    y' = 5x^4/(-5y^4)
    y' = x^4/(-y^4)

    y'' = [-y^4 * d/dx(x^4) - x^4 * d/dx(-y^4)]/(y^8)
    y'' = [-4y^4 * x^3 + 4x^4y^3 * y']/(y^8)
    substitute y'
    y'' = [-4y^4 * x^3 + 4x^4 * y^3 * (x^4/(-y^4))]/(y^8)
    y'' = [-4y^4 * x^3 - 4x^8/(y)]/(y^8)
    multiply by y
    y'' = -4[y^5 * x^3 + x^8)]/y^9
    substitute y^5 = 5 - x^5
    y'' = -4[(5 - x^5) * x^3 + x^8]/y^9
    y'' = -4[5x^3 - x^8 + x^8]/y^9
    y'' = -20x^3/y^9

    Sorry, I haven't quite had time to try LaTeX yet....
     
  6. May 8, 2007 #5
    Thank you very much Kuno.. great help.
     
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