# How do you implicitly differentiate -y=x-sqrt(x^3y-y) ?

May 23, 2018

$y ' = \frac{2 \sqrt{{x}^{3} y - y} - 3 {x}^{2} y}{{x}^{3} - 1 - 2 \sqrt{{x}^{3} y - y}}$

#### Explanation:

Differentiating with respect to $x$:
$- y ' = 1 - \frac{1}{2} {\left({x}^{3} y - y\right)}^{- \frac{1}{2}} \left(3 {x}^{2} y + {x}^{3} y ' - y '\right)$
Multiplying by $2 \sqrt{{x}^{2} y - y}$
${x}^{3} y ' - y ' - 2 y ' \sqrt{{x}^{3} y - y} = 2 \sqrt{{x}^{3} y - y} - 3 {x}^{2} y$
and this is
$y ' \left({x}^{3} - 1 - 2 \sqrt{{x}^{3} y - y}\right) = 2 \sqrt{{x}^{3} y - y} - 3 {x}^{2} y$
so
$y ' = \frac{2 \sqrt{{x}^{3} y - y} - 3 {x}^{2} y}{{x}^{3} - 1 - 2 \sqrt{{x}^{3} y - y}}$
if the denominator is not equal zero