#### Answer

$f'(a) = \dfrac{5}{(a+3)^2}$

#### Work Step by Step

Given $f(t) = \dfrac{2t+1}{t+3}$
$f'(a) = \lim\limits_{t \to a} \dfrac{f(t) - f(a)}{t - a} = \lim\limits_{t \to a} \dfrac{\frac{2t+1}{t+3} - \frac{2a+1}{a+3}}{t - a} = \lim\limits_{t \to a} \dfrac{\frac{(2t+1)(a+3) - (2a+1)(t+3)}{(t+3)(a+3)}}{t - a} = \lim\limits_{t \to a} \dfrac{(2t+1)(a+3) - (2a+1)(t+3)}{(t+3)(a+3)(t-a)} = \lim\limits_{t \to a} \dfrac{2at+a+6t+3 - (2at+t+6a+3)}{(t+3)(a+3)(t-a)} = \lim\limits_{t \to a} \dfrac{2at+a+6t+3 - 2at-t-6a-3}{(t+3)(a+3)(t-a)} = \lim\limits_{t \to a} \dfrac{a+6t-t-6a}{(t+3)(a+3)(t-a)} = \lim\limits_{t \to a} \dfrac{5t-5a}{(t+3)(a+3)(t-a)} = \lim\limits_{t \to a} \dfrac{5(t-a)}{(t+3)(a+3)(t-a)} = \lim\limits_{t \to a} \dfrac{5}{(t+3)(a+3)} = \dfrac{5}{(a+3)(a+3)} = \dfrac{5}{(a+3)^2}\longrightarrow f'(a) = \dfrac{5}{(a+3)^2}$