## University Calculus: Early Transcendentals (3rd Edition)

$s'=3(15t-1)^{-4}$
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, $s'=3(15t-1)^{-4}$