# How do you implicitly differentiate 4x^2+sqrt(2x+xy)+y=3?

Jul 13, 2018

$y ' = - \frac{16 x \sqrt{2 x + x y} + 2 + y}{x + 2 \sqrt{2 x + x y}}$

#### Explanation:

Assuming that $y = y \left(x\right)$ we get by the chain rule

$8 x + \frac{1}{2} {\left(2 x + x y\right)}^{- \frac{1}{2}} \left(2 + y + x y '\right) + y ' = 0$
multiplying by $2 \sqrt{2 x + x y}$ we get

$16 x \sqrt{2 x + x y} + 2 + y + x y ' + 2 y ' \sqrt{2 x + x y} = 0$ isolating $y '$

$16 x \sqrt{2 x + x y} + 2 + y + y ' \left(x + 2 \sqrt{2 x + x y}\right) = 0$ so we get

$y ' = - \frac{16 x \sqrt{2 x + x y} + 2 + y}{x + 2 \sqrt{2 x + x y}}$