Answer
$$r'\left( t \right) = \frac{{2\sqrt t + 1}}{{2\ln 3\sqrt t \left( {t + \sqrt t } \right)}}$$
Work Step by Step
$$\eqalign{
& r\left( t \right) = {\log _3}\left( {t + \sqrt t } \right) \cr
& {\text{Use logarithmic properties}} \cr
& r\left( t \right) = \frac{{\ln \left( {t + \sqrt t } \right)}}{{\ln 3}} \cr
& {\text{Differentiate}} \cr
& r'\left( t \right) = \frac{d}{{dt}}\left[ {\frac{{\ln \left( {t + \sqrt t } \right)}}{{\ln 3}}} \right] \cr
& r'\left( t \right) = \frac{1}{{\ln 3}}\frac{d}{{dt}}\left[ {\ln \left( {t + \sqrt t } \right)} \right] \cr
& {\text{Recall that }}\frac{d}{{dx}}\left[ {\ln u} \right] = \frac{{u'}}{u} \cr
& r'\left( t \right) = \frac{1}{{\ln 3}}\left( {\frac{{1 + \frac{1}{{2\sqrt t }}}}{{t + \sqrt t }}} \right) \cr
& {\text{Simplifying}} \cr
& r'\left( t \right) = \frac{1}{{\ln 3}}\left( {\frac{{2\sqrt t + 1}}{{2\sqrt t \left( {t + \sqrt t } \right)}}} \right) \cr
& r'\left( t \right) = \frac{{2\sqrt t + 1}}{{2\ln 3\sqrt t \left( {t + \sqrt t } \right)}} \cr} $$
