Answer
$$\frac{{ds}}{{dt}} = 8{\cos ^3}\left( {1 - 2t} \right)\sin \left( {1 - 2t} \right)$$
Work Step by Step
$$\eqalign{
& s = {\cos ^4}\left( {1 - 2t} \right) \cr
& {\text{differentiate with respect to }}t \cr
& \frac{{ds}}{{dt}} = \frac{d}{{dt}}\left[ {{{\cos }^4}\left( {1 - 2t} \right)} \right] \cr
& {\text{use the chain rule}} \cr
& \frac{{ds}}{{dt}} = 4{\cos ^3}\left( {1 - 2t} \right)\frac{d}{{dt}}\left[ {\cos \left( {1 - 2t} \right)} \right] \cr
& \frac{{ds}}{{dt}} = 4{\cos ^3}\left( {1 - 2t} \right)\left( { - \sin \left( {1 - 2t} \right)} \right)\frac{d}{{dt}}\left[ {1 - 2t} \right] \cr
& {\text{find the derivative and simplify}} \cr
& \frac{{ds}}{{dt}} = 4{\cos ^3}\left( {1 - 2t} \right)\left( { - \sin \left( {1 - 2t} \right)} \right)\left( { - 2} \right) \cr
& \frac{{ds}}{{dt}} = 8{\cos ^3}\left( {1 - 2t} \right)\sin \left( {1 - 2t} \right) \cr} $$
