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Differentiation question

  1. Jan 4, 2008 #1
    "the equation of a curve is x^2y=x-6 (x^2y is x squared times y)

    The normal of the curve at the point where x=-2 meets the x axis at p.

    Find the coordinates of P"

    The method I used was to find y by rearranging the given equation and differentiate it.
    This gave me f ' (x)= 1+12x^-3

    Then I subbed x=-2 in so that I could find the gradient of the tangent at that point. It worked out as -1/2, so then the gradient of the normal would be 2.

    After finding the equation of the line and making y=0, I got the wrong answer: x=-1 and y=0

    apparently the answer is (3/2, 0)

    any help would be appreciated
  2. jcsd
  3. Jan 4, 2008 #2
    your derivative is wrong
    get it right , you will have your answer.
  4. Jan 4, 2008 #3


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    If you rearrange the equation to y = (x - 6)/x^2, the derivative is (12-x)/x^3.
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