# How do you use implicit differentiation to find dy/dx given x^3+3x^2y+y^3=8?

Nov 5, 2016

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

#### Explanation:

differentiate $\textcolor{b l u e}{\text{implicitly with respect to x}}$

Use the $\textcolor{b l u e}{\text{product rule"" on the term }} 3 {x}^{2} y$

$\Rightarrow 3 {x}^{2} + 3 {x}^{2.} \frac{\mathrm{dy}}{\mathrm{dx}} + 6 x y + 3 {y}^{2.} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{3 {x}^{2} + 6 x y}{3 {x}^{2} + 3 {y}^{2}}$

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

$= - \frac{{x}^{2} + 2 x y}{{x}^{2} + {y}^{2}}$