Answer
$s'=\frac{1}{2t^{1/2}(1+t^{1/2})^2}$
Work Step by Step
Take the derivative of the equation using the Quotient Rule. Then apply Chain Rule:
$s'=\frac{(1+t^{1/2})(\frac{1}{2}t^{-1/2})-(t^{1/2})(\frac{1}{2}t^{-1/2})}{(1+t^{1/2})^2}$
$=\frac{\frac{1}{2}t^{-1/2}+\frac{1}{2}(1)-\frac{1}{2}(1)}{(1+t^{1/2})^2}$
$=\frac{1}{2t^{1/2}(1+t^{1/2})^2}$
