Answer
For $\cos (xy) =1+\sin y$ find $\frac{dy}{dx}$ by implicit differentiation
$\frac{dy}{dx}=\frac{y \sin(xy)}{-x\sin (xy) -\cos y}$
Work Step by Step
Differentiate both sides with respect to $x$ using the chain rule and the trig rules
$(-\sin (xy))(y+x\frac{dy}{dx})=(\cos(y))(\frac{dy}{dx})$
Isolate $\frac{dy}{dx}$
$-y\sin (xy)-x \sin(xy)\frac{dy}{dx}-\frac{dy}{dx}\cos y=0$
$\frac{dy}{dx}(-x\sin (xy) -\cos y)=y \sin(xy)$
$\frac{dy}{dx}=\frac{y \sin(xy)}{-x\sin (xy) -\cos y}$
