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Implicit physics question

  1. Nov 23, 2004 #1
    taking derivitive of 2xy^2+xy=y

    split up to using product rule

    [tex] 2x----2 [/tex]
    [tex] y^2---2y\frac{dy}{dx} [/tex]


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

    [tex] \frac{dy}{dx}= \frac{-2y^2-y}{4xy+x-1} [/tex]

    i am trying to figure out the slope of the equation when y-1

    am i doing all this for nothing?
    Last edited: Nov 23, 2004
  2. jcsd
  3. Nov 23, 2004 #2
    Do you mean when y=1?
    If so then all you have to do is plug back in y in the orignal equation and solve for x. And then plug in both x and y into your final differential equation. I think that the final slope will then be -9/2.
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