# Question #b46eb

Mar 18, 2016

$1. \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y - 2 x}{2 y - x}$
$2. \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\sin \frac{y}{x} ^ 2}{\left[\cos \frac{y}{x} - \frac{1}{y} ^ 2\right]}$

#### Explanation:

$1. \frac{d}{\mathrm{dx}} \left[{x}^{2} + {y}^{2} = x y + 3\right]$
$2 x + 2 y \frac{\mathrm{dy}}{\mathrm{dx}} = x \frac{\mathrm{dy}}{\mathrm{dx}} + y$
$2 y \frac{\mathrm{dy}}{\mathrm{dx}} - x \frac{\mathrm{dy}}{\mathrm{dx}} = y - 2 x$
$\frac{\mathrm{dy}}{\mathrm{dx}} \left(2 y - x\right) = y - 2 x$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y - 2 x}{2 y - x}$

$2. \sin \frac{y}{x} + \frac{1}{y} = 1$
$\frac{d}{\mathrm{dx}} \left[{x}^{-} 1 \sin y + {y}^{-} 1 = 1\right]$
$\frac{1}{x} \cos y \frac{\mathrm{dy}}{\mathrm{dx}} - \sin \frac{y}{x} ^ 2 - \frac{1}{y} ^ 2 \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
$\frac{1}{x} \cos y \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{1}{y} ^ 2 \frac{\mathrm{dy}}{\mathrm{dx}} = \sin \frac{y}{x} ^ 2$
$\frac{\mathrm{dy}}{\mathrm{dx}} \left[\cos \frac{y}{x} - \frac{1}{y} ^ 2\right] = \sin \frac{y}{x} ^ 2$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\sin \frac{y}{x} ^ 2}{\left[\cos \frac{y}{x} - \frac{1}{y} ^ 2\right]}$