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Implicit Differentiation - Tangent Line & Horizontal Tangents

  1. Mar 9, 2009 #1
    1. The problem statement, all variables and given/known data
    2181.jpg

    Find an equation of the tangent line to this curve at the point (1, -2).

    2. Relevant equations



    3. The attempt at a solution

    2y' = 3x^2+6x
    y' = 3x^2+6x
    y'=3/2x^2+3x

    y+2=3(x-1)
    y+2=3x-3
    y=3x-5
     
    Last edited: Mar 9, 2009
  2. jcsd
  3. Mar 9, 2009 #2

    lanedance

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    Hi Pondera

    where did the 2 go?
    where did the power of x go?
    I think the arithmetic needs a little work
     
  4. Mar 9, 2009 #3
    Lane, I divided 3x^2+6x by two and got y'=3/2x^2+3x.

    By power of x, I take it you mean power of 2? I neglected to put that in here, but it is on my scratch paper, I appologize.

    I don't see where the arithmetic is flawed? I certainly believe that that is likely not the equation/form that I need, but I believe I worked what I figured to be correct out correctly. Can you be more specific?
     
  5. Mar 9, 2009 #4

    lanedance

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    ok I don't really understand what you are trying to do, just picked up some misssing parts as discussed

    the question is clipped off when I veiw it
     
  6. Mar 9, 2009 #5

    lanedance

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    Also the implict derivative w.r.t. x of y2 is not 2y', it will be:

    [tex]\frac{d}{dx} y^2 = 2y \frac{dy}{dx} [/tex]

    horizontal derivatives where y' = 0
     
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