Answer
$y'=\cos{x}\cot^2{x}.$
Work Step by Step
$y=f(x)+g(x)\rightarrow f(x)=-\csc{x}$; $g(x)=-\sin{x}$
By Theorem 2.9: $f'(x)$ $=(-1)(-\csc{x}\cot{x})$
$=\csc{x}\cot{x}$
$g'(x)=\dfrac{d}{dx}(-\sin{x})=-\cos{x}$
$y'=f'(x)+g'(x)=\csc{x}\cot{x}-\cos{x}=\cos{x}\cot^2{x}.$
