# Math Help - implicit differentiation- where did i make a mistake

1. ## implicit differentiation- where did i make a mistake

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

i worked it all out but the answer im getting is not an option. could someone show me where i made a mistake? im 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!

2. Originally Posted by LexiRae
find d^2y/dx^2 at (1,1) for:
x^2y + x^2 -y^2 = 1

i worked it all out but the answer im getting is not an option. could someone show me where i made a mistake? im 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!
I'm struggling to read all that. Your expression for dy/dx is correct and the value at (1, 1) is correct. I suggest multiplying the top and bottom by -1 so that

$\frac{dy}{dx} = \frac{2xy + 2x}{2y - x^2}$.

Then $\frac{d^2y}{dx^2} = \frac{\left(2y + 2x \frac{dy}{dx} + 2\right) (2y - x^2) - \left(2 \frac{dy}{dx} - 2x \right) (2xy + 2x)}{(2y - x^2)^2}$.

Now substitute x = 1, y = 1 and $\frac{dy}{dx} = 4$. I get -12.