Question

2/2)-2sin(y)" align="absmiddle" class="latex-formula">

Answer 1

the assumption being that "y" is encapsulating a function in "x" terms.

$y+2=\cfrac{x^2}{2}-2sin(y)\implies \stackrel{\textit{taking the derivative to both sides}}{\cfrac{dy}{dx}+0=\cfrac{1}{2}(2x)\underset{chain~rule}{-2cos(y)\cdot \cfrac{dy}{dx}}} \\\\\\ \cfrac{dy}{dx}=x-2cos(y)\cfrac{dy}{dx}\implies \cfrac{dy}{dx}+2cos(y)\cfrac{dy}{dx}=x \\\\\\ \cfrac{dy}{dx}\left[ 1+2cos(y) \right]=x\implies \cfrac{dy}{dx}=\cfrac{x}{1+2cos(y)}$