Answer
$f'(a) = 6a^2+1$
Work Step by Step
Given $f(t) = 2t^3+t$
$f'(a) = \lim\limits_{t \to a} \dfrac{f(t)-f(a)}{t-a} = \lim\limits_{t \to a} \dfrac{2t^3+t-(2a^3+a)}{t-a} = \lim\limits_{t \to a} \dfrac{2t^3+t-2a^3-a}{t-a} = \lim\limits_{t \to a} \dfrac{2t^3-2a^3+t-a}{t-a} = \lim\limits_{t \to a} \dfrac{2(t^3-a^3)+(t-a)}{t-a} = \lim\limits_{t \to a} \dfrac{2(t-a)(t^2+t a+a^2)+(t-a)}{t-a} = \lim\limits_{t \to a} \dfrac{(t-a)[2(t^2+t a+a^2)+1]}{t-a} = \lim\limits_{t \to a} [2(t^2+t a+a^2)+1] = 2(a^2 + aa+ a^2)+1 = 2(a^2+a^2+a^2)+1= 2(3a^2)+1=6a^2+1 \longrightarrow f'(a)=6a^2+1$
