Answer
$y'=8cos^3(1-2t) \sin(1-2t)$
Work Step by Step
Given that $s=cos^4(1-2t)$
Apply derivative rules of differentiation:
$f(x)=p'(x)q(x)+p(x)q'(x)$
$y'=\dfrac{d}{dx}[ cos^4(1-2t)]$
$= 4cos^3(1-2t)(-sin (1-2t))\dfrac{d}{dx}(1-2t)$
$=4cos^3(1-2t)(-sin (1-2t))(-2)$
Hence, $y'=8cos^3(1-2t) \sin(1-2t)$
