# How do you implicitly differentiate -4=(x^2+y)^2-y^2x ?

Jun 21, 2016

$\frac{{y}^{2} - 4 {x}^{3} - 4 x y}{2 \left({x}^{2} + y - x y\right)}$

#### Explanation:

$\frac{d}{\mathrm{dx}} \left\{- 4 = {\left({x}^{2} + y\right)}^{2} - {y}^{2} x\right\}$

$\setminus \implies 0 = \setminus \textcolor{red}{2 \left({x}^{2} + y\right) \left(2 x + y '\right)} - \setminus \textcolor{b l u e}{2 y y ' x - {y}^{2}}$

red bit is chain rule, blue bit is product rule with a bit of chain...

$\setminus \implies 0 = 4 x \left({x}^{2} + y\right) + 2 y ' \left({x}^{2} + y - x y\right) - {y}^{2}$

$\setminus \implies y ' = \frac{{y}^{2} - 4 x \left({x}^{2} + y\right)}{2 \left({x}^{2} + y - x y\right)}$

$= \frac{{y}^{2} - 4 {x}^{3} - 4 x y}{2 \left({x}^{2} + y - x y\right)}$