Answer
$y = \frac{1}{4}x + \frac{1}{2}$
Work Step by Step
$$\eqalign{
& y = \frac{{1 + x}}{{1 + {e^x}}}{\text{ at the point }}\left( {0,\frac{1}{2}} \right) \cr
& {\text{Differentiate }}y{\text{ to calculate the slope at the point }}\left( {0,\frac{1}{2}} \right) \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\frac{{1 + x}}{{1 + {e^x}}}} \right] \cr
& {\text{Use the quotient rule}} \cr
& \frac{{dy}}{{dx}} = \frac{{\left( {1 + {e^x}} \right)\frac{d}{{dx}}\left[ {1 + x} \right] - \left( {1 + x} \right)\frac{d}{{dx}}\left[ {1 + {e^x}} \right]}}{{{{\left( {1 + {e^x}} \right)}^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{{\left( {1 + {e^x}} \right)\left( 1 \right) - \left( {1 + x} \right)\left( {{e^x}} \right)}}{{{{\left( {1 + {e^x}} \right)}^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{{1 + {e^x} - {e^x} - x{e^x}}}{{{{\left( {1 + {e^x}} \right)}^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{{1 - x{e^x}}}{{{{\left( {1 + {e^x}} \right)}^2}}} \cr
& {\text{Find }}m{\text{ at }}\left( {0,\frac{1}{2}} \right) \cr
& m = {\left. {\frac{{dy}}{{dx}}} \right|_{x = 0}} = \frac{{1 - \left( 0 \right){e^0}}}{{{{\left( {1 + {e^0}} \right)}^2}}} \cr
& m = \frac{1}{4} \cr
& \cr
& {\text{Use the Point}} - {\text{Slope Form of the Equation of a Line}} \cr
& y - {y_1} = m\left( {x - {x_1}} \right) \cr
& \underbrace {\left( {0,\frac{1}{2}} \right)}_{\left( {{x_1},{y_1}} \right)} \to x = 0{\text{ and }}{y_1} = \frac{1}{2} \cr
& {\text{Therefore}} \cr
& y - \frac{1}{2} = \frac{1}{4}\left( {x - 0} \right) \cr
& {\text{Simplify}} \cr
& y - \frac{1}{2} = \frac{1}{4}x \cr
& y = \frac{1}{4}x + \frac{1}{2} \cr} $$
