## Thomas' Calculus 13th Edition

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