# How do you find dy/dx for the function y=sin(2x+4y)?

Mar 9, 2015

You have an Implicit Function in which $y$ is function of $x$ so you have to derive it as well.
You get:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \left[\cos \left(2 x + 4 y\right)\right] \cdot \left[2 + 4 \frac{\mathrm{dy}}{\mathrm{dx}}\right]$
where you use the Chain Rule on the $\sin$ function.

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 \cos \left(2 x + 4 y\right) + 4 \cos \left(2 x + 4 y\right) \frac{\mathrm{dy}}{\mathrm{dx}}$
Collecting $\frac{\mathrm{dy}}{\mathrm{dx}}$:
$\frac{\mathrm{dy}}{\mathrm{dx}} \left[1 - 4 \cos \left(2 x + 4 y\right)\right] = 2 \cos \left(2 x + 4 y\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 \cos \left(2 x + 4 y\right)}{1 - 4 \cos \left(2 x + 4 y\right)}$

Hope it helps