Answer
$f'(x)=-\csc{x}(\csc^2{x}+\cot^2{x}).$
Work Step by Step
$f(x)=\dfrac{\cot{x}}{\sin{x}}=\cot{x}\csc{x}.$
Product Rule $(f’(x)=(u(x)(v(x))’=u’(x)v(x)+u(x)v’(x))$
$u(x)=\cot{x} ;u’(x)=-\csc^2{x} $.
$v(x)=\csc{x} ;v’(x)=-\csc{x}\cot{x} $.
$f'(x)=(-\csc^2{x})(\csc{x})+(-\csc{x}\cot{x})(\cot{x})$
$=-\csc{x}(\csc^2{x}+\cot^2{x})$.
