Given that x^6+2xy+y^2=4

find the value of d^2y/dx^2 (second derivative) at the point (1, 1).

I have worked it out and my ans is -7.5. I would like to know if this ans is correct.

Thanks.

x^6 + 2xy + y^2 = 4

6x^5 + 2(y + xy ') + 2y*y ' = 0

6x^5 + 2y + 2xy ' + 2y*y ' = 0

2xy ' + 2y*y ' = -6x^5 - 2y

y ' = [-6x^5 - 2y] / [2x + 2y] = [-3x^5 - y] / [x + y]
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y " = [(-15x^4 - y ')(x + y) - (-3x^5 - y)(1 + y ')] / (x + y)^2

y " = [(-15x^4 - [-3x^5 - y] / [x + y])(x + y) - (-3x^5 - y)(1 + [-3x^5 - y] / [x + y]] / (x + y)^2

y " = [-15x^4(x + y) - [-3x^5 - y]^2 / [x + y]) ] / (x+y)^2

y " = [-15x^4(x+y)^2 - (3x^5 + y)^2] / (x+y)^3

y "(1,1) = (-15*4 - 16) / 8 = -76 / 8 = -19/2

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Apparently I missed something above SOMEWHERE !

Used calculator to solve for y and differentiate twice. One case gives 15/2,
the other -15/2. You are correct.