How do we differentiate something like this? Chain rule should be used here, so the first term will be cos(x/y), but then I don't understand how to break up this (x/y) function?
$$\frac{d}{dx}\left[\sin(\dfrac{x}{y})\right]=\frac{d}{dx}\left[ x+y \right]$$
$$\cos(\dfrac{x}{y}) \cdot \dfrac{y-xy'}{y^2}=1+y'$$
$$\cos(\dfrac{x}{y})y - \cos(\dfrac{x}{y})xy' = y^2+y^2y'$$
$$\cos(\dfrac{x}{y})y - y^2 = \cos(\dfrac{x}{y})xy' + y^2y'$$
$$\cos(\dfrac{x}{y})y - y^2 = y' \left( \cos(\dfrac{x}{y})x + y^2 \right) $$
$$y'=\dfrac{\cos(\dfrac{x}{y})y - y^2}{\cos(\dfrac{x}{y})x + y^2 }$$
$\displaystyle \frac{dy}{dx} = \frac{-f_{x}}{f_{y}}$ where $\displaystyle f_{x}, f_{y}$ are the partial derivatives of x and y respectively. If you don't know what partial derivatives are, then for the x partial. treat y as though it is a constant. and similarly for the y partial, treat x as it is a constant.
I get $\displaystyle \frac{dy}{dx} = \frac{y}{x}$