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

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Answer 1 · 2016-10-16T19:26:17

$\frac{d^{2} y}{{d x}^{2}} = - \frac{1}{y^{3}}$ $(d^2y)/dx^2= -1/y^3$

$- 8 y (\frac{d y}{d x}) = 8 x$ $-8y(dy/dx) = 8x$

$\frac{d y}{d x} = \frac{- x}{y}$ $dy/dx = (-x)/y$

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

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

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

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

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

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

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

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

$\frac{d^{2} y}{{d x}^{2}} = - \frac{1}{y^{3}}$ $(d^2y)/dx^2= -1/y^3$

How do you find (d^2y)/(dx^2)d2ydx2(d^2y)/(dx^2) for -4y^2+4=4x^2−4y2+4=4x2-4y^2+4=4x^2?