Answer
$$\frac{{dy}}{{dt}} = ky\left( {1 - y} \right)$$
Work Step by Step
$$\eqalign{
& y = \frac{1}{{1 + b{e^{ - kt}}}} \cr
& {\text{Rewrite using exponential rules}} \cr
& y = {\left( {1 + b{e^{ - kt}}} \right)^{ - 1}} \cr
& {\text{Differentiate both sides with respect to }}t \cr
& \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {{{\left( {1 + b{e^{ - kt}}} \right)}^{ - 1}}} \right] \cr
& \frac{{dy}}{{dt}} = - {\left( {1 + b{e^{ - kt}}} \right)^{ - 2}}\frac{d}{{dt}}\left[ {1 + b{e^{ - kt}}} \right] \cr
& \frac{{dy}}{{dt}} = - {\left( {1 + b{e^{ - kt}}} \right)^{ - 2}}\left( { - kb{e^{ - kt}}} \right) \cr
& \frac{{dy}}{{dt}} = \frac{{kb{e^{ - kt}}}}{{{{\left( {1 + b{e^{ - kt}}} \right)}^2}}} \cr
& {\text{Rewrite}} \cr
& \frac{{dy}}{{dt}} = \frac{k}{{\left( {1 + b{e^{ - kt}}} \right)}}\left( {\frac{{b{e^{ - kt}}}}{{1 + b{e^{ - kt}}}}} \right) \cr
& {\text{Add and subtract }}1 \cr
& \frac{{dy}}{{dt}} = \frac{k}{{\left( {1 + b{e^{ - kt}}} \right)}}\left( {\frac{{1 + b{e^{ - kt}} - 1}}{{1 + b{e^{ - kt}}}}} \right) \cr
& \frac{{dy}}{{dt}} = \frac{k}{{\left( {1 + b{e^{ - kt}}} \right)}}\left( {\frac{{1 + b{e^{ - kt}}}}{{1 + b{e^{ - kt}}}} - \frac{1}{{1 + b{e^{ - kt}}}}} \right) \cr
& \frac{{dy}}{{dt}} = \frac{k}{{\left( {1 + b{e^{ - kt}}} \right)}}\left( {1 - \frac{1}{{1 + b{e^{ - kt}}}}} \right) \cr
& {\text{Where }}y = \frac{1}{{1 + b{e^{ - kt}}}} \cr
& \frac{{dy}}{{dt}} = ky\left( {1 - y} \right) \cr} $$
