Answer
$A'\left( p \right) = 12{p^2} + 3$
Work Step by Step
$$\eqalign{
& A\left( p \right) = 4{p^3} + 3p \cr
& {\text{The domain of the function is }}\left( { - \infty ,\infty } \right) \cr
& {\text{Using the definition of derivative}} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} \cr
& {\text{Therefore}}{\text{,}} \cr
& A'\left( p \right) = \mathop {\lim }\limits_{h \to 0} \frac{{4{{\left( {p + h} \right)}^3} + 3\left( {p + h} \right) - \left[ {4{p^3} + 3p} \right]}}{h} \cr
& {\text{Expand and simplify}} \cr
& A'\left( p \right) = \mathop {\lim }\limits_{h \to 0} \frac{{4\left( {{p^3} + 3{p^2}h + 3p{h^2} + {h^3}} \right) + 3\left( {p + h} \right) - 4{p^3} - 3p}}{h} \cr
& A'\left( p \right) = \mathop {\lim }\limits_{h \to 0} \frac{{4{p^3} + 12{p^2}h + 12p{h^2} + 4{h^3} + 3p + 3h - 4{p^3} - 3p}}{h} \cr
& A'\left( p \right) = \mathop {\lim }\limits_{h \to 0} \frac{{12{p^2}h + 12p{h^2} + 4{h^3} + 3h}}{h} \cr
& A'\left( p \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {12{p^2} + 12ph + 4{h^2} + 3} \right)h}}{h} \cr
& A'\left( p \right) = \mathop {\lim }\limits_{h \to 0} \left( {12{p^2} + 12ph + 4{h^2} + 3} \right) \cr
& {\text{Evaluate the limit when }}h \to 0 \cr
& A'\left( p \right) = 12{p^2} + 12p\left( 0 \right) + 4{\left( 0 \right)^2} + 3 \cr
& A'\left( p \right) = 12{p^2} + 3 \cr
& {\text{The domain of the derivative of the function is }}\left( { - \infty ,\infty } \right) \cr} $$
