Answer
$$\frac{{dp}}{{dq}} = \frac{1}{6}q + \frac{1}{{6{q^3}}} + \frac{1}{{{q^5}}}{\text{ and }}\frac{{{d^2}p}}{{d{q^2}}} = \frac{1}{6} - \frac{1}{{2{q^4}}} - \frac{5}{{{q^6}}}$$
Work Step by Step
$$\eqalign{
& p = \left( {\frac{{{q^2} + 3}}{{12q}}} \right)\left( {\frac{{{q^4} - 1}}{{{q^3}}}} \right) \cr
& {\text{Use the distributive property}} \cr
& p = \frac{{\left( {{q^2} + 3} \right)\left( {{q^4} - 1} \right)}}{{12q\left( {{q^3}} \right)}} \cr
& p = \frac{{{q^6} - {q^2} + 3{q^4} - 3}}{{12{q^4}}} \cr
& p = \frac{{{q^6}}}{{12{q^4}}} - \frac{{{q^2}}}{{12{q^4}}} + \frac{{3{q^4}}}{{12{q^4}}} - \frac{3}{{12{q^4}}} \cr
& p = \frac{{{q^2}}}{{12}} - \frac{1}{{12{q^2}}} + \frac{1}{4} - \frac{1}{{4{q^4}}} \cr
& p = \frac{{{q^2}}}{{12}} - \frac{1}{{12}}{q^{ - 2}} + \frac{1}{4} - \frac{1}{4}{q^{ - 4}} \cr
& \cr
& {\text{Find the first derivative}} \cr
& \frac{{dp}}{{dq}} = \frac{d}{{dq}}\left[ {\frac{{{q^2}}}{{12}} - \frac{1}{{12}}{q^{ - 2}} + \frac{1}{4} - \frac{1}{4}{q^{ - 4}}} \right] \cr
& {\text{use }}\frac{d}{{dq}}\left[ {{q^n}} \right] = n{q^{n - 1}}{\text{ and }}\frac{d}{{dq}}\left[ c \right] = 0,{\text{ }}c{\text{ is a constant}} \cr
& \frac{{dp}}{{dq}} = \frac{1}{{12}}\left( {2q} \right) - \frac{1}{{12}}\left( { - 2{q^{ - 3}}} \right) + 0 - \frac{1}{4}\left( { - 4{q^{ - 5}}} \right) \cr
& \frac{{dp}}{{dq}} = \frac{1}{6}q + \frac{1}{6}{q^{ - 3}} + {q^{ - 5}} \cr
& \frac{{dp}}{{dq}} = \frac{1}{6}q + \frac{1}{{6{q^3}}} + \frac{1}{{{q^5}}} \cr
& \cr
& {\text{Find the second derivative}} \cr
& \frac{{{d^2}p}}{{d{q^2}}} = \frac{d}{{dq}}\left[ {\frac{1}{6}q + \frac{1}{6}{q^{ - 3}} + {q^{ - 5}}} \right] \cr
& {\text{use }}\frac{d}{{dq}}\left[ {{q^n}} \right] = n{q^{n - 1}}{\text{ and }}\frac{d}{{dq}}\left[ c \right] = 0,{\text{ }}c{\text{ is a constant}} \cr
& \frac{{{d^2}p}}{{d{q^2}}} = \frac{1}{6}\left( 1 \right) + \frac{1}{6}\left( { - 3{q^{ - 4}}} \right) - 5{q^{ - 6}} \cr
& \frac{{{d^2}p}}{{d{q^2}}} = \frac{1}{6} - \frac{1}{2}{q^{ - 4}} - 5{q^{ - 6}} \cr
& \frac{{{d^2}p}}{{d{q^2}}} = \frac{1}{6} - \frac{1}{{2{q^4}}} - \frac{5}{{{q^6}}} \cr} $$
