# How do you find (dy)/(dx) given root3x+root3(y^4)=2?

Aug 16, 2016

Start by writing the roots as rational exponents:

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

Differentiate each term using the power rule:

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

$\frac{1}{3} {x}^{- \frac{2}{3}} + \frac{4}{3} {y}^{\frac{1}{3}} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$\frac{1}{3 {x}^{\frac{2}{3}}} + \frac{4}{3} {y}^{\frac{1}{3}} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$\frac{4}{3} {y}^{\frac{1}{3}} \frac{\mathrm{dy}}{\mathrm{dx}} = 0 - \frac{1}{3 {x}^{\frac{2}{3}}}$

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

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

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

If you want it in radical form instead of rational exponent form:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{4 \sqrt[3]{{x}^{2}} \sqrt[3]{y}}$

Hopefully this helps!