Answer
$$\frac{d}{dx}\sin \frac{y}{x}= \left(\frac{1}{x}\frac{dy}{dx}-\frac{y}{x^2}\right)\cos\frac{y}{x}.$$
Work Step by Step
Recall the quotient rule: $(\dfrac{u}{v})'=\dfrac{vu'-uv'}{v^2}$
Recall that $(\sin x)'=\cos x$.
Using the quotient and chain rules, we have
$$\frac{d}{dx}\sin \frac{y}{x}= \frac{x\frac{dy}{dx}-y}{x^2}\cos\frac{y}{x}= \left(\frac{1}{x}\frac{dy}{dx}-\frac{y}{x^2}\right)\cos\frac{y}{x}.$$
