Answer
$f'\left( x \right) = \frac{1}{{\sqrt {{e^{2x}} - 1} }}$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {\sec ^{ - 1}}\left( {{e^x}} \right) \cr
& {\text{Differentiate}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{{\sec }^{ - 1}}\left( {{e^x}} \right)} \right] \cr
& {\text{Use the Derivatives of Inverse Trigonometric Functions }}\left( {p223} \right) \cr
& {\text{and the chain rule}}{\text{. }}\frac{d}{{dx}}\left[ {{{\sec }^{ - 1}}u} \right] = \frac{1}{{u\sqrt {{u^2} - 1} }}\frac{{du}}{{dx}},\;u = {e^x} \cr
& f'\left( x \right) = \frac{1}{{{e^x}\sqrt {{{\left( {{e^x}} \right)}^2} - 1} }}\frac{d}{{dx}}\left[ {{e^x}} \right] \cr
& {\text{Therefore}}{\text{,}} \cr
& f'\left( x \right) = \frac{1}{{{e^x}\sqrt {{e^{2x}} - 1} }}\left( {{e^x}} \right) \cr
& f'\left( x \right) = \frac{1}{{\sqrt {{e^{2x}} - 1} }} \cr} $$
