# Find dy/ddx for xy² - sin(x+2y)= 2x?

Aug 20, 2015

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

#### Explanation:

Starting from the equation

$x {y}^{2} - \sin \left(x + 2 y\right) = 2 x$

We differentiate both sides of the equation, using $\frac{d}{\mathrm{dx}}$

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