Answer
$$g'\left( x \right) = - \cos x - \cot x\csc x$$
Work Step by Step
$$\eqalign{
& g\left( x \right) = \left( {\cos x} \right)\left( {\cot x} \right) \cr
& {\text{Differentiate}} \cr
& g'\left( x \right) = \frac{d}{{dx}}\left[ {\left( {\cos x} \right)\left( {\cot x} \right)} \right] \cr
& {\text{By the product rule}} \cr
& g'\left( x \right) = \cot x\left[ {\cos x} \right]' + \cos x\left[ {\cot x} \right]' \cr
& g'\left( x \right) = \cot x\left( { - \sin x} \right) + \cos x\left( { - {{\csc }^2}x} \right) \cr
& {\text{Simplifying}} \cr
& g'\left( x \right) = \left( {\frac{{\cos x}}{{\sin x}}} \right)\left( { - \sin x} \right) + \cos x\left( { - \frac{1}{{{{\sin }^2}x}}} \right) \cr
& g'\left( x \right) = - \cos x - \cot x\csc x \cr} $$
