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