## Thomas' Calculus 13th Edition

$$\frac{{dy}}{{dt}} = - \left( {3{t^2} + 6t + 2} \right)$$
\eqalign{ & y = \frac{1}{{t\left( {t + 1} \right)\left( {t + 2} \right)}} \cr & {\text{Take the natural log of both sides:}} \cr & {\text{Use the properties of logarithms}} \cr & \ln y = \ln \left( {\frac{1}{{t\left( {t + 1} \right)\left( {t + 2} \right)}}} \right) \cr & {\text{quotient rule:}} \cr & \ln y = \ln \left( 1 \right) - \ln t\left( {t + 1} \right)\left( {t + 2} \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}} \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}} = - \left( {3{t^2} + 6t + 2} \right) \cr}