## Calculus (3rd Edition)

$\frac{dy}{dx}=\frac{1-\cos(x+y)}{\cos (x+y)+\sin y}$
Implicitly differentiate $\sin (x+y)=x+\cos y$ with respect to $x$: $[\cos (x+y)](1+\frac{dy}{dx})=1+(-\sin y\times \frac{dy}{dx})$ Or $\cos(x+y)+\cos (x+y)\frac{dy}{dx}=1-\sin y \times\frac{dy}{dx}$ $\implies cos(x+y)\frac{dy}{dx}+\sin y\frac{dy}{dx}$$=1-\cos (x+y)$ Or $\frac{dy}{dx}=\frac{1-\cos(x+y)}{\cos (x+y)+\sin y}$