How do you find dy/dx by implicit differentiation given xy^3=y+x?

Feb 21, 2017

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

Explanation:

differentiate each term on both sides $\textcolor{b l u e}{\text{implicitly with respect to x}}$

$\text{Note differentiate " xy^3" using "color(blue)"product rule}$

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

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

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

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