Answer
$$\frac{{dy}}{{dx}} = - \frac{1}{{\sqrt {{e^{2x}} - 1} }}$$
Work Step by Step
$$\eqalign{
& y = {\csc ^{ - 1}}\left( {{e^x}} \right) \cr
& {\text{Differentiate with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {{{\csc }^{ - 1}}{e^x}} \right) \cr
& {\text{Use the formula of differentiation }}\,\,\frac{d}{{dx}}\left[ {{{\csc }^{ - 1}}u} \right] = - \frac{1}{{\left| u \right|\sqrt {{u^2} - 1} }}\frac{{du}}{{dx}} \cr
& \frac{{dy}}{{dx}} = - \frac{1}{{\left| {{e^x}} \right|\sqrt {{{\left( {{e^x}} \right)}^2} - 1} }}\frac{d}{{dx}}\left[ {{e^x}} \right] \cr
& \frac{{dy}}{{dx}} = - \frac{1}{{\left| {{e^x}} \right|\sqrt {{{\left( {{e^x}} \right)}^2} - 1} }}\left( {{e^x}} \right) \cr
& {\text{Simplifying}} \cr
& \frac{{dy}}{{dx}} = - \frac{{{e^x}}}{{{e^x}\sqrt {{e^{2x}} - 1} }} \cr
& \frac{{dy}}{{dx}} = - \frac{1}{{\sqrt {{e^{2x}} - 1} }} \cr} $$
