Implicit differentiation - messed up somewhere

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meredith
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Homework Statement


find d^2y/dx^2 at (1,1) for:
x^2y + x^2 -y^2 = 1




Homework Equations


none



The Attempt at a Solution


i worked it all out but the answer I am getting is not an option. could someone show me where i made a mistake? I am not asking you to do the problem for me, just fix my error. this is what i did:
first i found dy/dx:

x^2(dy/dx) + y(2x) +2x -2y(dy/dx)=0
(x^2 - 2y)(dy/dx) = -2xy - 2x
dy/dx = (-2xy-2x)/(x^2-2y)
when i put (1,1) in, i got 4

then to find the second derivative i used the quotient rule:
d^2y/dy^2 = [(x^2-2y) (-2x)(dy/dx) +y(-2) -2] - ((-2xy-2x)(2x-2)(dy/dx)]/(x^2-2y)^2

then i just put the values in. for the x's and y's i put in 1, and for dy/dx i put in 4. i ended up with:
[(-1 x -2 x 4 + -2 -2) - (-2 -2) x (2-2(4) ] / 1 = 28 ? but this answer is incorrect. what did i do wrong?


Thanks so much!
 
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It's not necessary to solve for dy/dx- just use implicit differentiation again.

You are correct that x^2(dy/dx) + y(2x) +2x -2y(dy/dx)= 0.

Now, (x^2(dy/dx)'= x^2 d^2y/dx^2+ 2x dy/dx, (2xy)'= 2xy'+ 2y, and (2y dy/dx)'= 2y d^2y/dx^2+ 2 (dy/dx)^2. Put those together.