## Calculus 8th Edition

For $\cos (xy) =1+\sin y$ find $\frac{dy}{dx}$ by implicit differentiation $\frac{dy}{dx}=\frac{y \sin(xy)}{-x\sin (xy) -\cos y}$
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 $\frac{dy}{dx}$ $-y\sin (xy)-x \sin(xy)\frac{dy}{dx}-\frac{dy}{dx}\cos y=0$ $\frac{dy}{dx}(-x\sin (xy) -\cos y)=y \sin(xy)$ $\frac{dy}{dx}=\frac{y \sin(xy)}{-x\sin (xy) -\cos y}$