Answer
$\dfrac{1}{(2t+1)^{2}}$
Work Step by Step
Consider $s(t) = \dfrac{t}{2t+1}$
We will use the derivative definition, such that
$s'(t) = \lim\limits_{h \to 0}\dfrac{\dfrac{(t+h)}{2(t+h)+1} - \dfrac{t}{(2t+1)}}{h}= \lim\limits_{h \to 0}[\dfrac{h}{h(2t+1)^{2}}]=\dfrac{1}{(2t+1)^{2}}$
