Answer
$$f'\left( s \right) = s{\left( {{s^2} - 1} \right)^{3/2}}\left( {8{s^3} - 3s + 25} \right)$$
Work Step by Step
$$\eqalign{
& f\left( s \right) = {\left( {{s^2} - 1} \right)^{5/2}}\left( {{s^3} + 5} \right) \cr
& {\text{Differentiating}} \cr
& f'\left( s \right) = \frac{d}{{ds}}\left[ {{{\left( {{s^2} - 1} \right)}^{5/2}}\left( {{s^3} + 5} \right)} \right] \cr
& {\text{By the product rule}} \cr
& f'\left( s \right) = \left( {{s^3} + 5} \right)\frac{d}{{ds}}\left[ {{{\left( {{s^2} - 1} \right)}^{5/2}}} \right] + {\left( {{s^2} - 1} \right)^{5/2}}\frac{d}{{ds}}\left[ {{s^3} + 5} \right] \cr
& f'\left( s \right) = \left( {{s^3} + 5} \right)\left( {\frac{5}{2}} \right){\left( {{s^2} - 1} \right)^{5/2 - 1}}\left( {2s} \right) + {\left( {{s^2} - 1} \right)^{5/2}}\left( {3{s^2}} \right) \cr
& {\text{Multiply and simplify}} \cr
& f'\left( s \right) = 5s\left( {{s^3} + 5} \right){\left( {{s^2} - 1} \right)^{3/2}} + 3{s^2}{\left( {{s^2} - 1} \right)^{5/2}} \cr
& {\text{Factoring}} \cr
& f'\left( s \right) = s{\left( {{s^2} - 1} \right)^{3/2}}\left[ {5\left( {{s^3} + 5} \right) + 3s\left( {{s^2} - 1} \right)} \right] \cr
& f'\left( s \right) = s{\left( {{s^2} - 1} \right)^{3/2}}\left( {5{s^3} + 25 + 3{s^3} - 3s} \right) \cr
& f'\left( s \right) = s{\left( {{s^2} - 1} \right)^{3/2}}\left( {8{s^3} - 3s + 25} \right) \cr} $$
