Answer
$$y' = \frac{{ab}}{{{e^x}{{\left( {1 + b{e^{ - x}}} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& y = \frac{a}{{1 + b{e^{ - x}}}} \cr
& {\text{write with negative exponents}} \cr
& y = a{\left( {1 + b{e^{ - x}}} \right)^{ - 1}} \cr
& {\text{find the derivative by the chain rule}} \cr
& y' = - a{\left( {1 + b{e^{ - x}}} \right)^{ - 2}}\left( {1 + b{e^{ - x}}} \right)' \cr
& y' = - a{\left( {1 + b{e^{ - x}}} \right)^{ - 2}}\left( { - b{e^{ - x}}} \right) \cr
& {\text{simplifying}} \cr
& y' = ab{e^{ - x}}{\left( {1 + b{e^{ - x}}} \right)^{ - 2}} \cr
& {\text{write with positive exponents}} \cr
& y' = \frac{{ab}}{{{e^x}{{\left( {1 + b{e^{ - x}}} \right)}^2}}} \cr} $$
