Answer
$$f''\left( x \right) = 2{\csc ^2}x\cot x$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \cot x \cr
& {\text{Differentiate}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\cot x} \right] \cr
& {\text{Using the basic rules of trigonometric differentiation}} \cr
& f'\left( x \right) = - {\csc ^2}x \cr
& {\text{Differentiate}} \cr
& f''\left( x \right) = \frac{d}{{dx}}\left[ { - {{\csc }^2}x} \right] \cr
& {\text{Use the general power rule for differentiation }} \cr
& f''\left( x \right) = - 2{\csc ^{2 - 1}}x\frac{d}{{dx}}\left[ {\csc x} \right] \cr
& f''\left( x \right) = - 2\csc x\left( { - \csc x\cot x} \right) \cr
& f''\left( x \right) = 2{\csc ^2}x\cot x \cr} $$
