Answer
$$\frac{{dy}}{{dt}} = 5{\left( {\frac{{\left( {t + 1} \right)\left( {t - 1} \right)}}{{\left( {t - 2} \right)\left( {t + 3} \right)}}} \right)^5}\left( {\frac{1}{{t + 1}} - \frac{1}{{t - 1}} - \frac{1}{{t - 2}} - \frac{1}{{t + 3}}} \right)$$
Work Step by Step
$$\eqalign{
& y = {\left( {\frac{{\left( {t + 1} \right)\left( {t - 1} \right)}}{{\left( {t - 2} \right)\left( {t + 3} \right)}}} \right)^5},\,\,\,\,\,t > 2 \cr
& {\text{Take the natural logarithm on both sides}} \cr
& \ln y = \ln {\left( {\frac{{\left( {t + 1} \right)\left( {t - 1} \right)}}{{\left( {t - 2} \right)\left( {t + 3} \right)}}} \right)^5} \cr
& {\text{use power property for logarithms}} \cr
& \ln y = 5\ln \left( {\frac{{\left( {t + 1} \right)\left( {t - 1} \right)}}{{\left( {t - 2} \right)\left( {t + 3} \right)}}} \right) \cr
& {\text{use quotient property for logarithms}} \cr
& \ln y = 5\ln \left( {\left( {t + 1} \right)\left( {t - 1} \right)} \right) - 5\ln \left( {\left( {t - 2} \right)\left( {t + 3} \right)} \right) \cr
& {\text{use product property for logarithms}} \cr
& \ln y = 5\ln \left( {t + 1} \right) + 5\ln \left( {t - 1} \right) - 5\ln \left( {t - 2} \right) - 5\ln \left( {t + 3} \right) \cr
& {\text{differentiate both sides with respect to }}t \cr
& \frac{1}{y}\frac{{dy}}{{dt}} = 5\left( {\frac{1}{{t + 1}}} \right) + 5\left( {\frac{1}{{t - 1}}} \right) - 5\left( {\frac{1}{{t - 2}}} \right) - 5\left( {\frac{1}{{t + 3}}} \right) \cr
& {\text{solve the equation for }}\frac{{dy}}{{dt}} \cr
& \frac{{dy}}{{dt}} = 5y\left( {\frac{1}{{t + 1}} - \frac{1}{{t - 1}} - \frac{1}{{t - 2}} - \frac{1}{{t + 3}}} \right) \cr
& {\text{replace }}y = {\left( {\frac{{\left( {t + 1} \right)\left( {t - 1} \right)}}{{\left( {t - 2} \right)\left( {t + 3} \right)}}} \right)^5} \cr
& \frac{{dy}}{{dt}} = 5{\left( {\frac{{\left( {t + 1} \right)\left( {t - 1} \right)}}{{\left( {t - 2} \right)\left( {t + 3} \right)}}} \right)^5}\left( {\frac{1}{{t + 1}} - \frac{1}{{t - 1}} - \frac{1}{{t - 2}} - \frac{1}{{t + 3}}} \right) \cr} $$
