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3x^2+2xy+y^2=2 then the value of dy/dx at x=1 is?

  • math -

    using implicit differentiation,

    6x + 2y + 2xy' + 2yy' = 0
    when x=1, y=-1, so plug them i:
    6 - 2 + 2y' - 2y' = 0
    Note that there is no solution, so the tangent to the curve is vertical. y' is not defined.

    Looking at dx/dy,
    6xx' + 2x + 2yx' + 2y = 0
    6x' + 2 - 2x' - 2 = 0
    4x' = 0
    x' = 0
    just as we suspected: dx/dy=0, so the slope is vertical.

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