Answer
$s'(t)=\dfrac{2t+5}{(4-5t-t^2)^2}.$
Work Step by Step
$s(t)=\dfrac{1}{4-5t-t^2}=(4-5t-t^2)^{-1}$
Using the Chain Rule:
$u=(4-5t-t^2)$; $\dfrac{du}{dt}=(-5-2t)$
$s(u)=u^{-1};\dfrac{d}{du}s(u)=-\dfrac{1}{u^2}$
$\dfrac{d}{dt}s(t)$ $=\dfrac{d}{du}s(u)\times\dfrac{du}{dt}$ $=(-\dfrac{1}{(4-5t-t^2)^2})(-2t-5)$
$=\dfrac{2t+5}{(4-5t-t^2)^2}.$
