Answer
$$\frac{{dy}}{{dx}} = - \frac{1}{{\sqrt {{e^{2t}} - 1} }}$$
Work Step by Step
$$\eqalign{
& y = {\csc ^{ - 1}}\left( {{e^t}} \right) \cr
& {\text{find the derivative of }}y{\text{ with respect to }}t \cr
& \frac{{dy}}{{dt}} = \frac{{d\left( {{{\csc }^{ - 1}}\left( {{e^t}} \right)} \right)}}{{dx}} \cr
& {\text{we can use the formula }}\cr
& \frac{{d\left( {{{\csc }^{ - 1}}u} \right)}}{{dt}} = - \frac{1}{{\left| u \right|\sqrt {{u^2} - 1} }}\frac{{du}}{{dt}}.\,\,\,\left( {{\text{see table 7}}{\text{.3}}} \right). \cr
& {\text{here }}u = {e^t};{\text{then}} \cr
& \frac{{dy}}{{dx}} = - \frac{1}{{\left| {{e^t}} \right|\sqrt {{{\left( {{e^t}} \right)}^2} - 1} }}\frac{{d\left( {{e^t}} \right)}}{{dt}} \cr
& \frac{{dy}}{{dx}} = - \frac{1}{{\left| {{e^t}} \right|\sqrt {{e^{2t}} - 1} }}\left( {{e^t}} \right) \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{dx}} = - \frac{1}{{{e^t}\sqrt {{e^{2t}} - 1} }}\left( {{e^t}} \right) \cr
& \frac{{dy}}{{dx}} = - \frac{1}{{\sqrt {{e^{2t}} - 1} }} \cr} $$
