How do you find $(d^2y)/(dx^2)$ for $-4y^2+4=4x^2$ ?

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Answer 1 · 2016-10-16T19:26:17

$(d^2y)/dx^2= -1/y^3$

$-8y(dy/dx) = 8x$

$dy/dx = (-x)/y$

$(d^2y)/dx^2 = d/dx(dy/dx)$

$(d^2y)/dx^2= (d((-x)/y))/dx$

$(d^2y)/dx^2= {-y - -x(dy/dx)}/y^2$

$(d^2y)/dx^2= {(-y^2)/y - -x((-x)/y)}/y^2$

$(d^2y)/dx^2= -{y^2/y + -x((-x)/y)}/y^2$

$(d^2y)/dx^2= -{y^2/y + x^2/y}/y^2$

$(d^2y)/dx^2= -{y^2 + x^2}/y^3$

From the original equation, $y^2 + x^2 = 1$ :

$(d^2y)/dx^2= -1/y^3$

How do you find (d^2y)/(dx^2)(d^2y)/(dx^2) for -4y^2+4=4x^2-4y^2+4=4x^2?