# How do you find dy/dx if cosy=e^(x+y)?

Feb 23, 2018

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

#### Explanation:

We want to $\frac{\mathrm{dy}}{\mathrm{dx}}$ when

$\cos \left(y\right) = {e}^{x + y}$

Differentiate the left side by the chain rule

$L H S = \frac{d}{\mathrm{dx}} \cos \left(y\right) = - \frac{\mathrm{dy}}{\mathrm{dx}} \sin \left(y\right)$

Differentiate the right side by the chain and product rule

$R H S = \frac{d}{\mathrm{dx}} {e}^{x + y} = \frac{d}{\mathrm{dx}} {e}^{x} {e}^{y} = {e}^{x} {e}^{y} + \frac{\mathrm{dy}}{\mathrm{dx}} {e}^{x} {e}^{y}$

These sides should equal

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

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} {e}^{x} {e}^{y} + \frac{\mathrm{dy}}{\mathrm{dx}} \sin \left(y\right) = - {e}^{x} {e}^{y}$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} \left({e}^{x} {e}^{y} + \sin \left(y\right)\right) = - {e}^{x} {e}^{y}$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{{e}^{x} {e}^{y}}{{e}^{x} {e}^{y} + \sin \left(y\right)}$