Answer
$s'\left( t \right) = \frac{{ - 4{t^3} - 9{t^2} + 24t + 6}}{{{{\left( {4t - 3} \right)}^5}}}$
Work Step by Step
$$\eqalign{
& s\left( t \right) = \frac{{{t^3} - 2t}}{{{{\left( {4t - 3} \right)}^4}}} \cr
& {\text{Differentiate}} \cr
& s'\left( t \right) = \frac{d}{{dt}}\left[ {\frac{{{t^3} - 2t}}{{{{\left( {4t - 3} \right)}^4}}}} \right] \cr
& {\text{Use the quotient rule rule for derivatives}} \cr
& s'\left( t \right) = \frac{{{{\left( {4t - 3} \right)}^4}\frac{d}{{dt}}\left[ {{t^3} - 2t} \right] - \left( {{t^3} - 2t} \right)\frac{d}{{dt}}\left[ {{{\left( {4t - 3} \right)}^4}} \right]}}{{{{\left[ {{{\left( {4t - 3} \right)}^4}} \right]}^2}}} \cr
& s'\left( t \right) = \frac{{{{\left( {4t - 3} \right)}^4}\left( {3{t^2} - 2} \right) - 4\left( {{t^3} - 2t} \right){{\left( {4t - 3} \right)}^3}\left( 4 \right)}}{{{{\left( {4t - 3} \right)}^8}}} \cr
& {\text{Simplifying}} \cr
& s'\left( t \right) = \frac{{\left( {4t - 3} \right)\left( {3{t^2} - 2} \right) - 4\left( {{t^3} - 2t} \right)\left( 4 \right)}}{{{{\left( {4t - 3} \right)}^5}}} \cr
& s'\left( t \right) = \frac{{12{t^3} - 8t - 9{t^2} + 6 - 16{t^3} + 32t}}{{{{\left( {4t - 3} \right)}^5}}} \cr
& s'\left( t \right) = \frac{{ - 4{t^3} - 9{t^2} + 24t + 6}}{{{{\left( {4t - 3} \right)}^5}}} \cr} $$
