# How do you find (dy)/(dx) given x^3+y+8x=2y^2?

Jun 1, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 {x}^{2} + 8}{4 y - 1}$

#### Explanation:

Use implicit differentiation and the shorthand $\frac{\mathrm{dy}}{\mathrm{dx}} = y '$

${x}^{3} + y + 8 x = 2 {y}^{2}$

$\frac{d}{\mathrm{dx}} \left({x}^{3} + y + 8 x\right) = \frac{d}{\mathrm{dx}} \left(2 {y}^{2}\right)$

$3 {x}^{2} + y ' + 8 = 4 y \cdot y '$

Now use algebra to solve for $y '$:

$3 {x}^{2} + 8 = 4 y \cdot y ' - y '$

$3 {x}^{2} + 8 = y ' \left(4 y - 1\right)$

$\frac{3 {x}^{2} + 8}{4 y - 1} = y '$

Therefore:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 {x}^{2} + 8}{4 y - 1}$

Jun 1, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 {x}^{2} + 8}{4 y - 1}$

#### Explanation:

Write the equation as:

${x}^{3} + 8 x = 2 {y}^{2} - y$

Differentiate both sides with respect to $x$:

$\frac{d}{\mathrm{dx}} \left({x}^{3} + 8 x\right) = \frac{d}{\mathrm{dx}} \left(2 {y}^{2} - y\right)$

$3 {x}^{2} + 8 = \left(4 y - 1\right) \frac{\mathrm{dy}}{\mathrm{dx}}$

So that:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 {x}^{2} + 8}{4 y - 1}$