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Logarithmic differentiation

  1. Sep 29, 2008 #1
    Another problem i'm not sure of :(

    find [tex]\frac{dy}{dx}[/tex] for the function xy[tex]^{2}[/tex] + x lnx = 4y

    my answer

    y[tex]^{2}[/tex] + x2y [tex]\frac{dy}{dx}[/tex] + lnx + x (1/x) [tex]\frac{dy}{dx}[/tex] = 4[tex]\frac{dy}{dx}[/tex]

    x2y [tex]\frac{dy}{dx}[/tex] + [tex]\frac{dy}{dx}[/tex] - 4[tex]\frac{dy}{dx}[/tex] = -y [tex]^{2}[/tex] - lnx

    [tex]\frac{dy}{dx}[/tex] ( x2y - 3) = -y[tex]^{2}[/tex] - lnx

    [tex]\frac{dy}{dx}[/tex] = [tex]\frac{-y ^{2} - lnx}{x2y - 3}[/tex]

    I'm not sure if this is the correct answer again any guidance is greatly appreciated!
    Thank you
     
  2. jcsd
  3. Sep 29, 2008 #2

    cristo

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    Staff Emeritus
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    Why does the fourth term here have a dy/dx in it? The derivative of xln(x) wrt x is ln(x)+x(1/x)
     
  4. Sep 29, 2008 #3
    so should the dy/dx be eliminated from the ln(x) + x(1/x) completely? leaving the resulting derivative equal to

    dy/dx = [tex]\frac{-y^{2} - lnx - 1}{ 2yx - 4}[/tex]
     
  5. Sep 29, 2008 #4

    Defennder

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    Yeah that looks correct.
     
  6. Sep 29, 2008 #5
    Thank you :D
     
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