Answer
$s'=3(15t-1)^{-4}$
Work Step by Step
Given that $s=\dfrac{-1}{15(15t-1)^3}$
Re-write the given equation as: $s=\dfrac{-1}{15}(15t-1)^{-3}$
Apply derivative rules of differentiation:
$f(x)=p'(x)q(x)+p(x)q'(x)$
$s'=\dfrac{d}{dx}[\dfrac{-1}{15}(15t-1)^{-3}]$
$=\dfrac{3}{15}(15t-1)^{-4}\dfrac{d}{dx}(15t-1)$
$=\dfrac{3}{15}(15t-1)^{-4}(15)$
$=\dfrac{1}{5}(15t-1)^{-4}(15)$
Hence, $s'=3(15t-1)^{-4}$
