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

##### 1 Answer
Oct 16, 2016

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{1}{y} ^ 3$

#### Explanation:

$- 8 y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 8 x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- x}{y}$

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{d}{\mathrm{dx}} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)$

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{d \left(\frac{- x}{y}\right)}{\mathrm{dx}}$

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

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{\frac{- {y}^{2}}{y} - - x \left(\frac{- x}{y}\right)}{y} ^ 2$

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

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

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{{y}^{2} + {x}^{2}}{y} ^ 3$

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

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{1}{y} ^ 3$