Answer
$s'=-2(\dfrac{4t}{t+1})^{-3}[\dfrac{4}{(t+1)^2}]$
Work Step by Step
Given that $s=(\dfrac{4t}{t+1})^{-2}$
Apply derivative rules of differentiation:
$f(x)/g(x)=\dfrac{p'(x)q(x)-p(x)q'(x)}{g(x))^2}$
$s'=\dfrac{d}{dx}[(\dfrac{4t}{t+1})^{-2}]$
$=-2(\dfrac{4t}{t+1})^{-3}\dfrac{d}{dx}(\dfrac{4t}{t+1})$
$=-2(\dfrac{4t}{t+1})^{-3}[\dfrac{4t+4-4t}{(t+1)^2}]$
Hence, $s'=-2(\dfrac{4t}{t+1})^{-3}[\dfrac{4}{(t+1)^2}]$
