# How do you implicitly differentiate sqrt(xy)= x - 2y?

Jun 1, 2015

First, write it as ${\left(x y\right)}^{\frac{1}{2}} = x - 2 y$ or ${x}^{\frac{1}{2}} {y}^{\frac{1}{2}} = x - 2 y$.

Next, differentiate both sides with respect to $x$, assuming that $y$ is a function of $x$. You'll need the Product Rule and the Chain Rule:

$\setminus \frac{1}{2} {x}^{- \frac{1}{2}} {y}^{\frac{1}{2}} + \setminus \frac{1}{2} {x}^{\frac{1}{2}} {y}^{- \frac{1}{2}} \setminus \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 1 - 2 \frac{\mathrm{dy}}{\mathrm{dx}}$.

Finally, solve this equation for $\frac{\mathrm{dy}}{\mathrm{dx}}$:

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \setminus \frac{1 - \setminus \frac{1}{2} {x}^{- \frac{1}{2}} {y}^{\frac{1}{2}}}{\setminus \frac{1}{2} {x}^{\frac{1}{2}} {y}^{- \frac{1}{2}} + 2} = \setminus \frac{1 - \setminus \frac{\setminus \sqrt{y}}{2 \setminus \sqrt{x}}}{\setminus \frac{\setminus \sqrt{x}}{2 \setminus \sqrt{y}} + 2} = \setminus \frac{2 \setminus \sqrt{x y} - y}{x + 4 \setminus \sqrt{x y}}$