If $\sin (x+y)=e^{x+y}-2$, then $\frac{d y}{d x}$ is equal to

x = 0 x = 2 x = −1 x = 3

x = −1

$\sin (x+y)=e^{x+y}-2$ differentiating wrt x $\cos(x + y)[1+\frac{dy}{dx}]=e^{x+y}[1+\frac{dy}{dx}]$ $⇒\frac{dy}{dx}=-1$