# How do you find the second derivative of y^2 - 16x = 0?

Nov 22, 2016

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{- 64}{{y}^{3}}$

#### Explanation:

${y}^{2} - 16 x = 0$

Implicitly differentiate to find first derivative:
$2 y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) - 16 = 0$
$\frac{\mathrm{dy}}{\mathrm{dx}} \left(2 y\right) = 16$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{16}{2 y}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{8}{y}$

Implicitly differentiate to find second derivative (use quotient rule ):
$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{\left(y\right) \left(0\right) - \left(8\right) \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}{{y}^{2}}$

Substitute in $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{8}{y}$
$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{- \left(8\right) \left(\frac{8}{y}\right)}{{y}^{2}}$
$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{\frac{- 64}{y}}{{y}^{2}}$
$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{- 64}{{y}^{3}}$

I am not sure if this answer can be simplified or not by substituting the value of y from the original equation...