Answer
$$\frac{{dy}}{{dx}} = - 2{x^3}{\csc ^2}\left( {\frac{1}{2}{x^4}} \right)$$
Work Step by Step
$$\eqalign{
& y = \cot \left( {\frac{1}{2}{x^4}} \right) \cr
& {\text{differentiate with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\cot \left( {\frac{1}{2}{x^4}} \right)} \right] \cr
& {\text{using the chain rule for }}{D_x}\left( {\cot u} \right) = - {\csc ^2}u \cdot {D_x}\left( u \right).{\text{ for this exercise }}u = \frac{1}{2}{x^4} \cr
& {\text{then}} \cr
& \frac{{dy}}{{dx}} = - {\csc ^2}\left( {\frac{1}{2}{x^4}} \right)\frac{d}{{dx}}\left[ {\frac{1}{2}{x^4}} \right] \cr
& {\text{solve the derivative}} \cr
& \frac{{dy}}{{dx}} = - {\csc ^2}\left( {\frac{1}{2}{x^4}} \right)\left( {2{x^3}} \right) \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{dx}} = - 2{x^3}{\csc ^2}\left( {\frac{1}{2}{x^4}} \right) \cr} $$
