Ive been banging my head on this one for 30 minutes..
xsin(y) = cos(x+y) , then find y'
$\displaystyle \frac{d}{dx} (x \sin y) = \frac{d}{dx} \cos(x+y)$
First, the left side using the product rule:
$\displaystyle \frac{d}{dx} (x \sin y) = \frac{d}{dx} (x) \cdot \sin y + x \cdot \frac{d}{dx} (\sin y) = \sin y + x \cos y \, \frac{dy}{dx}$
Now the other side:
$\displaystyle \frac{d}{dx} \cos(x+y) = -\sin(x+y) \cdot \frac{d}{dx} (x+y) = -\sin(x+y) \cdot \left(1+\frac{dy}{dx}\right)$
So we now have:
$\displaystyle \sin y + x \cos y \, \frac{dy}{dx} = -\sin(x+y) \cdot \left(1+\frac{dy}{dx}\right)$
At this point is it an algebra problem to solve for $\displaystyle \frac{dy}{dx}$. Can you finish this?