Answer
$$\frac{dy}{dx}=-\csc^2 x\cos(\cot x)$$
Work Step by Step
$$y=\sin(\cot x)$$ $$\frac{dy}{dx}=\frac{d(\sin(\cot x))}{dx}$$
Let $u=\cot x$ and $y=\sin u$. Then, according to Chain Rule, $$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$$ $$\frac{dy}{dx}=\frac{d(\sin u)}{du}\frac{d(\cot x)}{dx}$$ $$\frac{dy}{dx}=\cos u\times(-\csc^2 x)$$ $$\frac{dy}{dx}=-\csc^2 x\cos(\cot x)$$
