## Calculus 10th Edition

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