## Thomas' Calculus 13th Edition

$$\frac{{dy}}{{dt}} = \frac{1}{{2\sqrt 2 t\sqrt {\ln t} }}$$
\eqalign{ & y = \sqrt {\ln \sqrt t } \cr & {\text{rewrite the radicals}} \cr & y = {\left( {\ln {t^{1/2}}} \right)^{1/2}} \cr & {\text{use the logarithmic property }}\ln {t^n} = n\ln t \cr & y = {\left( {\frac{1}{2}\ln t} \right)^{1/2}} \cr & y = {\left( {\frac{1}{2}} \right)^{1/2}}{\left( {\ln t} \right)^{1/2}} \cr & {\text{find the derivative of }}y{\text{ with respect to }}t \cr & \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {{{\left( {\frac{1}{2}} \right)}^{1/2}}{{\left( {\ln t} \right)}^{1/2}}} \right] \cr & \frac{{dy}}{{dt}} = {\left( {\frac{1}{2}} \right)^{1/2}}\frac{d}{{dt}}\left[ {{{\left( {\ln t} \right)}^{1/2}}} \right] \cr & {\text{use the chain rule}} \cr & \frac{{dy}}{{dt}} = {\left( {\frac{1}{2}} \right)^{1/2}}\left( {\frac{1}{2}} \right){\left( {\ln t} \right)^{ - 1/2}}\frac{d}{{dt}}\left[ {\ln t} \right] \cr & {\text{then}} \cr & \frac{{dy}}{{dt}} = {\left( {\frac{1}{2}} \right)^{3/2}}{\left( {\ln t} \right)^{ - 1/2}}\left( {\frac{1}{t}} \right) \cr & {\text{simplifying, we get:}} \cr & \frac{{dy}}{{dt}} = \frac{1}{{2\sqrt 2 t\sqrt {\ln t} }} \cr & \frac{{dy}}{{dt}} = \frac{1}{{2\sqrt 2 t\sqrt {\ln t} }} \cr}