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Approximating slope of the derivative at a point

  1. Nov 29, 2007 #1
    1. The problem statement, all variables and given/known data
    The slope of the line tangent to the curve y^3 + x^2y^2 - 3x^3 = 9 at (1.5,2) is approximately


    2. Relevant equations
    Answer choices
    A) 0.39
    B) 3.2
    C) -11.45
    D) -2.29
    E) -3.2


    3. The attempt at a solution
    Well, I solved for the derivative, but I'm not sure if it was to any avail seeing as when I plug the point back in I do not get an answer choice.

    I got:

    dy/dx = (-2xy^(2) + 9x^2) / (3y + 2x^(2)y)

    Did I solve the derivative incorrectly? Or is plugging in the point not the correct course of action? Thank you again for all your help, I've been struggling with this problem a lot and any guidance/explanation would be great!
     
  2. jcsd
  3. Nov 29, 2007 #2
    Hello,

    It looks like you made a small error in your implicit differentiation-- go back and double check, see if you can spot it.

    Hope this helps.
     
  4. Dec 1, 2007 #3
    Yep, I think you simply dropped an exponent on the y from the denominator. I got

    [tex]\frac{dy}{dx}=\frac{9x^{2}-2xy^{2}}{3y^{2}+2x^{2}y}[/tex]

    Plugging in the numbers, I get .393. Hope that helps.
     
    Last edited: Dec 1, 2007
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