Answer
$$f'(a)=\frac{5}{(a+3)^2}$$
Work Step by Step
$$f(t)=\frac{2t+1}{t+3}$$
That means $$f(a)=\frac{2a+1}{a+3}$$
The derivative of $f(t)$ at number $a$ can be found as follows: $$f'(a)=\lim\limits_{t\to a}\frac{f(t)-f(a)}{t-a}$$
$$f'(a)=\lim\limits_{t\to a}\frac{\frac{2t+1}{t+3}-\frac{2a+1}{a+3}}{t-a}$$
$$f'(a)=\lim\limits_{t\to a}\frac{\frac{(2t+1)(a+3)-(2a+1)(t+3)}{(t+3)(a+3)}}{t-a}$$
$$f'(a)=\lim\limits_{t\to a}\frac{(2at+6t+a+3)-(2at+6a+t+3)}{(t+3)(a+3)(t-a)}$$
$$f'(a)=\lim\limits_{t\to a}\frac{5t-5a}{(t+3)(a+3)(t-a)}$$
$$f'(a)=\lim\limits_{t\to a}\frac{5}{(t+3)(a+3)}$$
$$f'(a)=\frac{5}{(a+3)(a+3)}$$
$$f'(a)=\frac{5}{(a+3)^2}$$
