## Thomas' Calculus 13th Edition

$$\frac{{dy}}{{dt}} = 3{t^2} + 6t + 2$$
\eqalign{ & y = t\left( {t + 1} \right)\left( {t + 2} \right) \cr & {\text{Take the natural logarithm of both sides:}} \cr & {\text{Use the properties of logarithms}} \cr & \ln y = \ln \left( {t\left( {t + 1} \right)\left( {t + 2} \right)} \right) \cr & {\text{product rule:}} \cr & \ln y = \ln \left( t \right) + \ln \left( {t + 1} \right) + \ln \left( {t + 2} \right) \cr & {\text{Take derivatives of both sides with respect to }}t \cr & \frac{1}{y}\frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {\left( {\ln t} \right)} \right] + \frac{d}{{dt}}\left[ {\ln \left( {t + 1} \right)} \right] + \frac{d}{{dt}}\left[ {\ln \left( {t + 2} \right)} \right] \cr & \frac{1}{y}\frac{{dy}}{{dt}} = \frac{1}{t} + \frac{1}{{t + 1}} + \frac{1}{{t + 2}} \cr & {\text{solve for }}\frac{{dy}}{{dt}} \cr & \frac{{dy}}{{dt}} = y\left( {\frac{1}{t} + \frac{1}{{t + 1}} + \frac{1}{{t + 2}}} \right) \cr & {\text{substitute }}t\left( {t + 1} \right)\left( {t + 2} \right){\text{ for }}y{\text{ }} \cr & \frac{{dy}}{{dt}} = t\left( {t + 1} \right)\left( {t + 2} \right)\left( {\frac{1}{t} + \frac{1}{{t + 1}} + \frac{1}{{t + 2}}} \right) \cr & \frac{{dy}}{{dt}} = t\left( {t + 1} \right)\left( {t + 2} \right)\left( {\frac{{{t^2} + 3t + 2 + {t^2} + 2t + {t^2} + t}}{{t\left( {t + 1} \right)\left( {t + 2} \right)}}} \right) \cr & \frac{{dy}}{{dt}} = 3{t^2} + 6t + 2 \cr}