# How do you find the derivative of x^3 + (x^2)(y) + y^3 = 9 at P(-1,2)?

Sep 25, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{13}$

#### Explanation:

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

$\text{differentiate "x^2y" using the "color(blue)"product rule}$

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

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

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

$\text{At } P \left(- 1 , 2\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{3 - 4}{1 + 12} = \frac{1}{13}$