Answer
$$ - \frac{1}{2}$$
Work Step by Step
$$\eqalign{
& \frac{d}{{dx}}{\left. {\left( {{{\tan }^{ - 1}}{e^{ - x}}} \right)} \right|_{x = 0}} \cr
& {\text{set }}y = {\tan ^{ - 1}}{e^{ - x}} \cr
& {\text{differentiate }}y \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}{e^{ - x}}} \right] \cr
& {\text{use }}\frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}u} \right] = \frac{1}{{1 + {u^2}}}\frac{{du}}{{dx}}.{\text{ consider }}u = {e^{ - x}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{1 + {{\left( {{e^{ - x}}} \right)}^2}}}\frac{d}{{dx}}\left[ {{e^{ - x}}} \right] \cr
& \frac{{dy}}{{dx}} = \frac{1}{{1 + {{\left( {{e^{ - x}}} \right)}^2}}}\left( { - {e^{ - x}}} \right) \cr
& \frac{{dy}}{{dx}} = - \frac{{{e^{ - x}}}}{{1 + {e^ - }^{2x}}} \cr
& {\text{substitute }}y = {\tan ^{ - 1}}{e^{ - x}} \cr
& \frac{d}{{dx}}\left( {{{\tan }^{ - 1}}{e^{ - x}}} \right) = - \frac{{{e^{ - x}}}}{{1 + {e^ - }^{2x}}} \cr
& {\text{evaluate at }}x = 0 \cr
& \frac{d}{{dx}}{\left. {\left( {{{\tan }^{ - 1}}{e^{ - x}}} \right)} \right|_{x = 0}} = - \frac{{{e^{ - 0}}}}{{1 + {e^ - }^{2\left( 0 \right)}}} \cr
& {\text{simplifying}} \cr
& \frac{d}{{dx}}{\left. {\left( {{{\tan }^{ - 1}}{e^{ - x}}} \right)} \right|_{x = 0}} = - \frac{1}{{1 + 1}} \cr
& \frac{d}{{dx}}{\left. {\left( {{{\tan }^{ - 1}}{e^{ - x}}} \right)} \right|_{x = 0}} = - \frac{1}{2} \cr} $$
