Answer
$$s'\left( t \right) = - \frac{{3{\pi ^2}}}{{{{\left( {3t + 4} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& s\left( t \right) = \frac{{{\pi ^2}}}{{3t + 4}} \cr
& {\text{Use }}\frac{1}{u} = {u^{ - 1}} \cr
& s\left( t \right) = {\pi ^2}{\left( {3t + 4} \right)^{ - 1}} \cr
& {\text{Calculate the derivative of the function}} \cr
& s'\left( t \right) = \frac{d}{{dt}}\left( {{\pi ^2}{{\left( {3t + 4} \right)}^{ - 1}}} \right) \cr
& s'\left( t \right) = {\pi ^2}\frac{d}{{dt}}\left( {{{\left( {3t + 4} \right)}^{ - 1}}} \right) \cr
& {\text{use the chain rule}} \cr
& s'\left( t \right) = {\pi ^2}\left( { - 1} \right){\left( {3t + 4} \right)^{ - 2}}\frac{d}{{dt}}\left( {3t + 4} \right) \cr
& s'\left( t \right) = {\pi ^2}\left( { - 1} \right){\left( {3t + 4} \right)^{ - 2}}\left( 3 \right) \cr
& s'\left( t \right) = - \frac{{3{\pi ^2}}}{{{{\left( {3t + 4} \right)}^2}}} \cr} $$
