# What is the implicit derivative of 1= ysqrt(xy)-y ?

Dec 13, 2015

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{y \sqrt{x y}}{x \left(3 \sqrt{x y} - 2\right)}$

#### Explanation:

$\frac{d}{\mathrm{dx}} \left[1 = y \sqrt{x y} - y\right]$

$0 = \sqrt{x y} \frac{\mathrm{dy}}{\mathrm{dx}} + y \frac{d}{\mathrm{dx}} \left[\sqrt{x y}\right] - \frac{\mathrm{dy}}{\mathrm{dx}}$

$0 = \sqrt{x y} \frac{\mathrm{dy}}{\mathrm{dx}} + y \left(\frac{1}{2 \sqrt{x y}}\right) \frac{d}{\mathrm{dx}} \left[x y\right] - \frac{\mathrm{dy}}{\mathrm{dx}}$

$0 = \sqrt{x y} \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{y}{2 \sqrt{x y}} \left(y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right) - \frac{\mathrm{dy}}{\mathrm{dx}}$

$0 = \sqrt{x y} \frac{\mathrm{dy}}{\mathrm{dx}} + {y}^{2} / \left(2 \sqrt{x y}\right) + \frac{x y}{2 \sqrt{x y}} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) - \frac{\mathrm{dy}}{\mathrm{dx}}$

$0 = \sqrt{x y} \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{y \sqrt{x y}}{2 x} + \frac{\sqrt{x y}}{2} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) - \frac{\mathrm{dy}}{\mathrm{dx}}$

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

$- \frac{y \sqrt{x y}}{2 x} = \frac{\mathrm{dy}}{\mathrm{dx}} \left(\frac{3 \sqrt{x y} - 2}{2}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{y \sqrt{x y}}{2 x} \left(\frac{2}{3 \sqrt{x y} - 2}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{y \sqrt{x y}}{x \left(3 \sqrt{x y} - 2\right)}$

Please ask with any questions! Above all, never forget to use the product rule.