Answer
$$s'\left( x \right) = \frac{{ - 28\left( {2x + 4} \right)}}{{{{\left( {3x - 1} \right)}^3}}}$$
Work Step by Step
$$\eqalign{
& s\left( x \right) = {\left( {\frac{{2x + 4}}{{3x - 1}}} \right)^2} \cr
& {\text{Use property of exponents}} \cr
& s\left( x \right) = \frac{{{{\left( {2x + 4} \right)}^2}}}{{{{\left( {3x - 1} \right)}^2}}} \cr
& {\text{Differentiate}} \cr
& s'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{{{\left( {2x + 4} \right)}^2}}}{{{{\left( {3x - 1} \right)}^2}}}} \right] \cr
& {\text{By the quotient rule and use the chain rule and the power rule }} \cr
& \frac{d}{{dx}}\left[ {{u^n}} \right] = n{u^{n - 1}}u' \cr
& s'\left( x \right) = \frac{{{{\left( {3x - 1} \right)}^2}\left( 2 \right)\left( {2x + 4} \right)\left( 2 \right) - {{\left( {2x + 4} \right)}^2}\left( 2 \right)\left( {3x - 1} \right)\left( 3 \right)}}{{{{\left( {3x - 1} \right)}^4}}} \cr
& {\text{Simplifying}} \cr
& s'\left( x \right) = \frac{{4\left( {2x + 4} \right){{\left( {3x - 1} \right)}^2} - 6{{\left( {2x + 4} \right)}^2}\left( {3x - 1} \right)}}{{{{\left( {3x - 1} \right)}^4}}} \cr
& s'\left( x \right) = \frac{{4\left( {2x + 4} \right)\left( {3x - 1} \right) - 6{{\left( {2x + 4} \right)}^2}}}{{{{\left( {3x - 1} \right)}^3}}} \cr
& s'\left( x \right) = \frac{{2\left( {2x + 4} \right)\left[ {2\left( {3x - 1} \right) - 3\left( {2x + 4} \right)} \right]}}{{{{\left( {3x - 1} \right)}^3}}} \cr
& s'\left( x \right) = \frac{{2\left( {2x + 4} \right)\left( {6x - 2 - 6x - 12} \right)}}{{{{\left( {3x - 1} \right)}^3}}} \cr
& s'\left( x \right) = \frac{{ - 28\left( {2x + 4} \right)}}{{{{\left( {3x - 1} \right)}^3}}} \cr} $$
