# How do you implicitly differentiate  x^3 + 3x^2 + y^3 = 8?

Apr 11, 2016

You can do it like this:

#### Explanation:

${x}^{3} + 3 {x}^{2} + {y}^{3} = 8$

Differentiating both sides with respect to $x$:

$D \left({x}^{3} + 3 {x}^{2} + {y}^{3}\right) = D \left(8\right)$

$\therefore 3 {x}^{2} + 6 x + 3 {y}^{2} y ' = 0$

$\therefore 3 {y}^{2} y ' = - \left(3 {x}^{2} + 6 x\right) = - 3 \left({x}^{2} + 2 x\right)$

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

$y ' = \frac{\left({x}^{2} + 2 x\right)}{{y}^{2}}$