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Find the equation of y^2=x(x-3)^2 of tangent line at (3,0)

  1. Feb 6, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the equation of y^2=x(x-3)^2 of tangent line at (3,0)

    2. Relevant equations
    Given above.
    I think implicit differentiation is involved or no since there is no xy's on the same side?


    3. The attempt at a solution
    Anyways...

    My attempt:

    2ydy/dx = x*2(x-3)*1
    2ydy/dx=2x(x-3)
    dy/dx= 2x(x-3)/2y
    dy/dx= 2x^2-6x/2y


    Doesn't seem right.
     
  2. jcsd
  3. Feb 6, 2013 #2
    Another problem" Determine (x,y) location(s) where the graph of y^4 = y^2-x^2 has horizontal tangents

    I got the answer dy/dx = -2x/4y^3-2y

    I don't know how to calculate x and y positions? I just found the implicit differentiation
     
  4. Feb 6, 2013 #3

    Mark44

    Staff: Mentor

    You have a mistake above. x * (x - 3)2 is a product, so the product rule is needed.

    After you use the product rule, you will need to use the chain rule.
     
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