# How do you find dy/dx by implicit differentiation for sinx + 2cos2y = 1?

Aug 30, 2015

$\frac{\mathrm{dy}}{\mathrm{dx}} = \cos \frac{x}{4 \sin \left(2 y\right)}$

#### Explanation:

Differentiate every term with respect to $x$

$\cos \left(x\right) - 4 \sin \left(2 y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

isolate the term containing $\frac{\mathrm{dy}}{\mathrm{dx}}$ by adding

$4 \sin \left(2 y\right) \frac{\mathrm{dy}}{\mathrm{dx}}$ to both sides

$4 \sin \left(2 y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = \cos \left(x\right)$

Divide both sides by $4 \sin \left(2 y\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \cos \frac{x}{4 \sin \left(2 y\right)}$