## Thomas' Calculus 13th Edition

$$\frac{{dy}}{{dt}} = - \frac{1}{2}\sqrt {\frac{1}{{t\left( {t + 1} \right)}}} \left( {\frac{{1 + 2t}}{{t\left( {t + 1} \right)}}} \right)$$
\eqalign{ & y = \sqrt {\frac{1}{{t\left( {t + 1} \right)}}} \cr & y = {\left( {\frac{1}{{t\left( {t + 1} \right)}}} \right)^{1/2}} \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)}}} \right)^{1/2}} \cr & {\text{power rule:}} \cr & \ln y = \frac{1}{2}\ln \left( {\frac{1}{{t\left( {t + 1} \right)}}} \right) \cr & {\text{quotient rule:}} \cr & \ln y = \frac{1}{2}\ln \left( 1 \right) - \frac{1}{2}\ln \left( {t\left( {t + 1} \right)} \right) \cr & {\text{product rule}} \cr & \ln y = - \frac{1}{2}\ln \left( t \right) - \frac{1}{2}\ln \left( {t + 1} \right) \cr & {\text{Take derivatives of both sides}} \cr & \frac{1}{y}\frac{{dy}}{{dt}} = - \frac{1}{2}\left( {\frac{1}{t}} \right) - \frac{1}{2}\left( {\frac{1}{{t + 1}}} \right) \cr & \frac{1}{y}\frac{{dy}}{{dt}} = - \frac{1}{2}\left( {\frac{1}{t} + \frac{1}{{t + 1}}} \right) \cr & {\text{solve for }}\frac{{dy}}{{dt}} \cr & \frac{{dy}}{{dt}} = - \frac{y}{2}\left( {\frac{{t + 1 + t}}{{t\left( {t + 1} \right)}}} \right) \cr & \frac{{dy}}{{dt}} = - \frac{y}{2}\left( {\frac{{1 + 2t}}{{t\left( {t + 1} \right)}}} \right) \cr & {\text{substitute }}\sqrt {\frac{1}{{t\left( {t + 1} \right)}}} {\text{ for }}y{\text{ }} \cr & \frac{{dy}}{{dt}} = - \frac{1}{2}\sqrt {\frac{1}{{t\left( {t + 1} \right)}}} \left( {\frac{{1 + 2t}}{{t\left( {t + 1} \right)}}} \right) \cr}