Answer
$s'=-2(\frac{4t}{t+1})^{-3}(\frac{4}{(t+1)^2})$
Work Step by Step
Take the derivative of the equation using Power Rule, Chain Rule, and Quotient Rule:
$s'=-2\times(\frac{4t}{t+1})^{-3}\times\frac{(t+1)(4)-(4t)(1)}{(t+1)^2}$
$=-2(\frac{4t}{t+1})^{-3}(\frac{4t+4-4t}{(t+1)^2})$
$=-2(\frac{4t}{t+1})^{-3}(\frac{4}{(t+1)^2})$
