# Question #43877

##### 1 Answer
Feb 25, 2018

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

#### Explanation:

Differentiate both sides with respect to $x$. This means that every time we differentiate a term containing $y$, we should end up with an instance of $\frac{\mathrm{dy}}{\mathrm{dx}}$.

$\frac{d}{\mathrm{dx}} \left(x \cos \left(y\right)\right) = \frac{d}{\mathrm{dx}} \left(\sin \left(x + y\right)\right)$

$\cos \left(y\right) \frac{d}{\mathrm{dx}} \left(x\right) + x \frac{d}{\mathrm{dx}} \cos \left(y\right) = \cos \left(x + y\right) \frac{d}{\mathrm{dx}} \left(x + y\right)$

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

Solve for $\frac{\mathrm{dy}}{\mathrm{dx}} :$

$- x \sin \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}} + \cos \left(y\right) = \cos \left(x + y\right) + \frac{\mathrm{dy}}{\mathrm{dx}} \cos \left(x + y\right)$ (Multiply out on the right side)

$- x \sin \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{\mathrm{dy}}{\mathrm{dx}} \cos \left(x + y\right) = \cos \left(x + y\right) - \cos \left(y\right)$ (Isolate all terms containing $\frac{\mathrm{dy}}{\mathrm{dx}}$ on the left side, move all other terms to the right)

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(- x \sin \left(y\right) - \cos \left(x + y\right)\right) = \cos \left(x + y\right) - \cos \left(y\right)$ (Factor out $\frac{\mathrm{dy}}{\mathrm{dx}}$)

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\cos \left(x + y\right) - \cos \left(y\right)}{- x \sin \left(y\right) - \cos \left(x + y\right)}$
(Divide both sides by $- x \sin \left(y\right) - \cos \left(x + y\right)$)