# How do you find (d^2y)/(dx^2) for 5=4x^3-4y^2?

Nov 4, 2016

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{12 x {y}^{2} - 9 {x}^{4}}{4 {y}^{3}}$

#### Explanation:

$5 = 4 {x}^{3} - 4 {y}^{2}$

$0 = 12 {x}^{2} - 8 y \frac{\mathrm{dy}}{\mathrm{dx}}$

$8 y \frac{\mathrm{dy}}{\mathrm{dx}} = 12 {x}^{2}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{12 {x}^{2}}{8 y} = \frac{3 {x}^{2}}{2 y}$

To find the second derivative use the quotient rule

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{2 y \cdot 6 x - 3 {x}^{2} \cdot 2 \frac{\mathrm{dy}}{\mathrm{dx}}}{4 {y}^{2}}$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{2 \left(6 x y - 3 {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}\right)}{4 {y}^{2}}$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{6 x y - 3 {x}^{2} \cdot \frac{3 {x}^{2}}{2 y}}{2 {y}^{2}}$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{6 x y - \frac{9 {x}^{4}}{2 y}}{2 {y}^{2}}$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{\frac{12 x {y}^{2} - 9 {x}^{4}}{2 y}}{2 {y}^{2}}$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \left(\frac{12 x {y}^{2} - 9 {x}^{4}}{2 y}\right) \cdot \left(\frac{1}{2 {y}^{2}}\right)$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{12 x {y}^{2} - 9 {x}^{4}}{4 {y}^{3}}$