# Thread:

1. ## 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

2. ## Re: Implicit Differentiation

Originally Posted by kspkido
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.

3. ## Re: Implicit Differentiation

You found 1 point, but there is another.

4. ## 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)

5. ## 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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