# How do you find dy/dx by implicit differentiation of sqrt(xy)=x-2y?

Dec 10, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{4}$

#### Explanation:

This is a homogeneous equation and no need for implicit differentiation.

Putting $y = \lambda x$ into $\sqrt{x y} = x - 2 y$ we have

$x \sqrt{\lambda} = x - 2 \lambda x$ or

$\left(\sqrt{\lambda} - 1 + 2 \lambda\right) x = 0$ but $x \ne 0$ so

$\sqrt{\lambda} - 1 + 2 {\left(\sqrt{\lambda}\right)}^{2} = 0$

so

$\sqrt{\lambda} = \frac{- 1 \pm \sqrt{1 + 8}}{4} = \frac{- 1 \pm 3}{4} = \frac{1}{2}$ so we have

$\frac{y}{x} = \lambda = \frac{1}{4}$ and finally

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{4}$