# How do you find dy/dx by implicit differentiation of (x+y)^3=x^3+y^3 and evaluate at point (-1,1)?

Apr 6, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2 x y + {y}^{2}}{{x}^{2} + 2 x y}$ and at $\left(- 1 , 1\right)$, $\frac{\mathrm{dy}}{\mathrm{dx}} = - 1$

#### Explanation:

As ${\left(x + y\right)}^{3} = {x}^{3} + {y}^{3}$, taking differential on both sides implicitly, we get

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

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

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

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

and at $\left(- 1 , 1\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2 \left(- 1\right) \times 1 + {1}^{2}}{{1}^{2} + 2 \left(- 1\right) \left(1\right)} = - \frac{- 1}{- 1} = - 1$