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Math Help - Implicit Differentiation

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    Implicit Differentiation

    I've tried differentiating this over again, and took me a while to get it. but Im not sure, my answer were (1/3, 4/3)

    heres the question
    At what point on the curve xy=(1-x-y)^2 is the tangent line parallel to the x-axis.

    Thank you very much
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    Re: Implicit Differentiation

    Quote Originally Posted by kspkido View Post
    I've tried differentiating this over again, and took me a while to get it. but Im not sure, my answer were (1/3, 4/3)

    heres the question
    At what point on the curve xy=(1-x-y)^2 is the tangent line parallel to the x-axis.

    Thank you very much
    Hint: If the tangent line is parallel to the x axis, then the gradient is 0. Therefore, you need to evaluate the point where dy/dx = 0.
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    Re: Implicit Differentiation

    You found 1 point, but there is another.
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    Re: Implicit Differentiation

    wow. okay thanks..but okay to tell you the truth, i didn't actually solve the problem i just saw the answer at the back of the book, I know the process differentiate it then set the dy/dx zero since parallel means same slope and x-axis is is horizontal meaning slope is zero.. but what i encountered is I don't really know how to get the rest of the point...i tried substituting the y to x in the dy/dx formula but i can never separate y from x... so there's 2 integers on dy/dx=-(y-2-2x)/(2+2x+4y)
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    MHF Contributor MarkFL's Avatar
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    Re: Implicit Differentiation

    Implicitly differentiating, you should find:

    \frac{dy}{dx}=\frac{2x+y-2}{2-x-2y}

    Equating this to zero implies:

    y=2(1-x)

    Substituting this into the original equation gives:

    -2x(x-1)=(x-1)^2

    (x-1)(3x-1)=0

    x=\frac{1}{3},1

    So the two points are: \left(\frac{1}{3},\frac{4}{3} \right),(1,0)
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