Answer
$$\frac{{dy}}{{dx}} = - 15{\cot ^4}x{\csc ^2}x$$
Work Step by Step
$$\eqalign{
& y = 3{\cot ^5}x \cr
& {\text{we can write the function as}} \cr
& y = 3{\left( {\cot x} \right)^5} \cr
& {\text{differentiate with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {3{{\left( {\cot x} \right)}^5}} \right] \cr
& \frac{{dy}}{{dx}} = 3\frac{d}{{dx}}\left[ {{{\left( {\cot x} \right)}^5}} \right] \cr
& {\text{use the general power rule for derivatives }}\frac{d}{{dx}}\left[ {{u^n}} \right] = n{u^{n - 1}}\frac{{du}}{{dx}}.{\text{ consider }}u = \cos x \cr
& then \cr
& \frac{{dy}}{{dx}} = 3\left( 5 \right){\left( {\cot x} \right)^{5 - 1}}\frac{d}{{dx}}\left[ {\cot x} \right] \cr
& {\text{use }}{D_x}\left( {\cot x} \right) = - {\csc ^2}x \cr
& \frac{{dy}}{{dx}} = 3\left( 5 \right){\left( {\cot x} \right)^4}\left( { - {{\csc }^2}x} \right) \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{dx}} = - 15{\cot ^4}x{\csc ^2}x \cr} $$
