Answer
$p' = \frac{-1}{2q^{2}}$
$p'' = \frac{1}{q^{3}}$
Work Step by Step
$p = \frac{q^{2}+3}{(q-1)^{3}+(q+1)^{3}}$
$p = \frac{q^{2}+3}{q^{3}-1-3q^{2}+3q+q^{3}+1+3q^{2}+3q}$
$p = \frac{q^{2}+3}{2q(q^{2}+3)}$
$p = \frac{1}{2q}$
$p' = \frac{-1}{2q^{2}}$
$p'' = \frac{1}{q^{3}}$