Answer
$$g'\left( t \right) = {t^2}{\left( {{t^4} + 5} \right)^{5/2}}\left( {17{t^4} + 15} \right)$$
Work Step by Step
$$\eqalign{
& g\left( t \right) = {t^3}{\left( {{t^4} + 5} \right)^{7/2}} \cr
& {\text{differentiate both sides}} \cr
& g'\left( t \right) = \frac{d}{{dt}}\left[ {{t^3}{{\left( {{t^4} + 5} \right)}^{7/2}}} \right] \cr
& {\text{use the product rule}} \cr
& g'\left( t \right) = {t^3}\frac{d}{{dt}}\left[ {{{\left( {{t^4} + 5} \right)}^{7/2}}} \right] + {\left( {{t^4} + 5} \right)^{7/2}}\frac{d}{{dt}}\left[ {{t^3}} \right] \cr
& {\text{use the chain rule }}\frac{d}{{dx}}\left[ {g{{\left( x \right)}^n}} \right] = ng{\left( x \right)^{n - 1}}\frac{d}{{dx}}\left[ {g'\left( x \right)} \right]{\text{ then}} \cr
& g'\left( t \right) = {t^3}\left( {\frac{7}{2}} \right){\left( {{t^4} + 5} \right)^{5/2}}\frac{d}{{dt}}\left[ {{t^4} + 5} \right] + {\left( {{t^4} + 5} \right)^{7/2}}\frac{d}{{dt}}\left[ {{t^3}} \right] \cr
& {\text{find derivatives}} \cr
& g'\left( t \right) = {t^3}\left( {\frac{7}{2}} \right){\left( {{t^4} + 5} \right)^{5/2}}\left( {4{t^3}} \right) + {\left( {{t^4} + 5} \right)^{7/2}}\left( {3{t^2}} \right) \cr
& {\text{simplifying}} \cr
& g'\left( t \right) = 14{t^6}{\left( {{t^4} + 5} \right)^{5/2}} + 3{t^2}{\left( {{t^4} + 5} \right)^{7/2}} \cr
& g'\left( t \right) = {t^2}{\left( {{t^4} + 5} \right)^{5/2}}\left[ {14{t^4} + 3\left( {{t^4} + 5} \right)} \right] \cr
& g'\left( t \right) = {t^2}{\left( {{t^4} + 5} \right)^{5/2}}\left( {14{t^4} + 3{t^4} + 15} \right) \cr
& g'\left( t \right) = {t^2}{\left( {{t^4} + 5} \right)^{5/2}}\left( {17{t^4} + 15} \right) \cr} $$
