## Thomas' Calculus 13th Edition

$6$
Use derivative formula; $f(x)= x^n \implies f'(x)= n x^{n-1}$ Consider $s(t) = 1-3t^{2}$ Thus, $s'(t) = 0 - 3(2)t^{2-1}= -6t$ and $s'(-1) = -6(-1)=6$