Answer
$s'=-5\csc^5(1-t+3t^2)cot(1-t+3t^2)(-1+6t)$
Work Step by Step
Given that $s=\csc^5(1-t+3t^2)$
Apply derivative rules of differentiation:
$f(x)=p'(x)q(x)+p(x)q'(x)$
$y'=\dfrac{d}{dx}[\csc^5(1-t+3t^2)]$
$=5\csc^4(1-t+3t^2)\dfrac{d}{dx}(\csc(1-t+3t^2))$
$=5\csc^4(1-t+3t^2)(-csc(1-t+3t^2))(1-t+3t^2)\dfrac{d}{dx}((1-t+3t^2))$
$=5\csc^4(1-t+3t^2)(-csc(1-t+3t^2))(1-t+3t^2)(-1+6t)$
Hence, $s'=-5\csc^5(1-t+3t^2)cot(1-t+3t^2)(-1+6t)$
