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March 29, 2017

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Find the equation of the tangent line to the curve x^(3)+3xy+y^(3)=1 at the point (1,0).

  • Calculus - ,

    Implicit derivative:

    3x^2 + 3x dy/dx + 3y + 3y^2 dy/dx = 0
    dy/dx(3x + 3y^2) = -3x^2 - 3y
    dy/dx = (-x^2 - y)/(x + y^2)

    at (1,0)
    dy/dx = -1 - 0)/(1 + 0) = -1

    let y = -x + b
    at (1,0)
    0 = -1 + b
    b = 1

    tangent equation at (1,0) is
    y = -x + 1

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