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Finding the derivative.

  1. Apr 8, 2008 #1
    1. The problem statement, all variables and given/known data

    Find [tex]\frac{dy}{dx}[/tex] if y = [tex]\sqrt{5u^2 -3}[/tex] and u = [tex]\frac{2x}{3x+1}[/tex]


    2. Relevant equations

    Chain Rule
    [tex]\frac{dy}{dx}[/tex] = [tex]\frac{dy}{du}[/tex] x [tex]\frac{du}{dx}[/tex]


    3. The attempt at a solution

    [tex]\frac{dy}{du}[/tex] = 5u(5u^2 - 3)^-1/2

    [tex]\frac{du}{dx}[/tex] = -6x(3x+1)^-2

    [tex]\frac{dy}{dx}[/tex] = 5u(5u^2 - 3)^-1/2 x -6x(3x+1)^-2
    = -30xu(5u^2 - 3)^-1/2 x (3x+1)^-2
    = [tex]\frac{-60x^2}{3x+1}[/tex](5([tex]\frac{2x}{3x+1}[/tex])^2 - 3)^-1/2 x (3x+1)^-2

    Is that the final simplified answer?
     
  2. jcsd
  3. Apr 8, 2008 #2

    G01

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    Check your work for du/dx. I think you may have an error. You should have a 2 in the numerator, not a -6x.
     
  4. Apr 8, 2008 #3
    u = [tex]\frac{2x}{3x+1}[/tex]
    u = (2x)(3x+1)^-1
    u' = -2x(3x+1)^-2 x (3)
    u' = -6x(3x+1)^-2

    Did I do something wrong?
     
  5. Apr 8, 2008 #4

    CompuChip

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    [tex]\frac{d}{dx}(f(x) g(x)) = \frac{df(x)}{dx} g(x) + f(x) \frac{dg(x)}{dx}[/tex]
    You have a term missing.
     
  6. Apr 8, 2008 #5
    u = (2x)(3x+1)^-1
    u' = 2(3x+1)^-1 + (2x)(-1)(3x+1)^2(3)
    = 2(3x+1)^-1 - 6x(3x+1)^-2

    This?
     
  7. Apr 8, 2008 #6

    G01

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    Yes, that's the correct du/dx.
     
  8. Apr 8, 2008 #7
    dy/dx = 5u(5u^2 - 3)^-1/2 x (2(3x+1)^-1 - 6x(3x+1)^-2)

    From here do I just sub in for u?
     
  9. Apr 8, 2008 #8

    G01

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    Yep. You got it.
     
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