Answer
$\frac{{dy}}{{dx}} = \frac{{{e^x}}}{{{{\left( {1 - {e^x}} \right)}^2}}}$
Work Step by Step
$$\eqalign{
& y = \frac{{{e^x}}}{{1 - {e^x}}} \cr
& {\text{Differentiating}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\frac{{{e^x}}}{{1 - {e^x}}}} \right] \cr
& {\text{Differentiate by using the formula }}\left( {\frac{f}{g}} \right)' = \frac{{gf' - fg'}}{{{g^2}}} \cr
& {\text{Let }}f = {e^x}{\text{ and }}g = 1 - {e^x},{\text{ then}} \cr
& \frac{{dy}}{{dx}} = \frac{{\left( {1 - {e^x}} \right)\left( {{e^x}} \right)' - {e^x}\left( {1 - {e^x}} \right)'}}{{{{\left( {1 - {e^x}} \right)}^2}}} \cr
& {\text{Compute the derivatives}} \cr
& \frac{{dy}}{{dx}} = \frac{{\left( {1 - {e^x}} \right){e^x} - {e^x}\left( { - {e^x}} \right)}}{{{{\left( {1 - {e^x}} \right)}^2}}} \cr
& {\text{Multiply and simplify}} \cr
& \frac{{dy}}{{dx}} = \frac{{{e^x} - {e^{2x}} + {e^{2x}}}}{{{{\left( {1 - {e^x}} \right)}^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{{{e^x}}}{{{{\left( {1 - {e^x}} \right)}^2}}} \cr} $$
