Answer
$$f'\left( t \right) = 8t + 1$$
Work Step by Step
$$\eqalign{
& f\left( t \right) = 4{t^2} + t \cr
& {\text{Using the definition of the derivative of a function}} \cr
& f'\left( t \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {t + h} \right) - f\left( t \right)}}{h} \cr
& f'\left( t \right) = \mathop {\lim }\limits_{h \to 0} \frac{{4{{\left( {t + h} \right)}^2} + \left( {t + h} \right) - \left( {4{t^2} + t} \right)}}{h} \cr
& {\text{Simplifying}} \cr
& f'\left( t \right) = \mathop {\lim }\limits_{h \to 0} \frac{{4\left( {{t^2} + 2th + {h^2}} \right) + \left( {t + h} \right) - \left( {4{t^2} + t} \right)}}{h} \cr
& f'\left( t \right) = \mathop {\lim }\limits_{h \to 0} \frac{{4{t^2} + 8th + 4{h^2} + t + h - 4{t^2} - t}}{h} \cr
& f'\left( t \right) = \mathop {\lim }\limits_{h \to 0} \frac{{8th + 4{h^2} + h}}{h} \cr
& f'\left( t \right) = \mathop {\lim }\limits_{h \to 0} \left( {8t + 4h + 1} \right) \cr
& {\text{Evaluating the limit}} \cr
& f'\left( t \right) = 8t + 4\left( 0 \right) + 1 \cr
& f'\left( t \right) = 8t + 1 \cr} $$
