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

Feb 22, 2017

$\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{2}{9 {y}^{2}}$ or $\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{2}{9 {\left({\left(\left(\frac{2}{3}\right) x - \left(\frac{2}{3}\right)\right)}^{\frac{1}{3}}\right)}^{2}}$

#### Explanation:

$9 {y}^{2} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 2$

Solve for $\frac{\mathrm{dy}}{\mathrm{dt}}$

$\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{2}{9 {y}^{2}}$

If you want, we can solve and plug in for y using the given equation and simplifying:

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

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

${y}^{3} = \left(\frac{2}{3}\right) x - \left(\frac{2}{3}\right)$

$y = {\left(\left(\frac{2}{3}\right) x - \left(\frac{2}{3}\right)\right)}^{\frac{1}{3}}$

Now plug this in for y and simplify.

$\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{2}{9 {\left({\left(\left(\frac{2}{3}\right) x - \left(\frac{2}{3}\right)\right)}^{\frac{1}{3}}\right)}^{2}}$

$\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{2}{9 \left({\left(\left(\frac{2}{3}\right) x - \left(\frac{2}{3}\right)\right)}^{\frac{2}{3}}\right)}$