Answer
$s'=3cot^2(\dfrac{2}{t})\csc^2(\dfrac{2}{t})\dfrac{2}{t^2}$
Work Step by Step
Given that $s=cot^3(\dfrac{2}{t})$
Re-write the given equation as:$s=cot^3(2t^{-1})$
Apply derivative rules of differentiation:
$f(x)=p'(x)q(x)+p(x)q'(x)$
$y'=\dfrac{d}{dx}[cot^3(2t^{-1})]$
$=3 cot^2(2t^{-1})(-csc^2(2t^{-1}))\dfrac{d}{dx}(2t^{-1})$
Hence, $s'=3cot^2(\dfrac{2}{t})\csc^2(\dfrac{2}{t})\dfrac{2}{t^2}$
