## Thomas' Calculus 13th Edition

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