Answer
$$\frac{{dy}}{{dx}} = \sec x$$
Work Step by Step
$$\eqalign{
& y = {\sinh ^{ - 1}}\left( {\tan x} \right) \cr
& {\text{Differentiate both sides}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{{\sinh }^{ - 1}}\left( {\tan x} \right)} \right] \cr
& {\text{Use Theorem 5}}{\text{.20 }}\frac{d}{{dx}}\left[ {{{\sinh }^{ - 1}}u} \right] = \frac{{u'}}{{\sqrt {{u^2} + 1} }} \cr
& \frac{{dy}}{{dx}} = \frac{{\frac{d}{{dx}}\left[ {\tan x} \right]}}{{\sqrt {{{\left( {\tan x} \right)}^2} + 1} }} \cr
& {\text{Computing derivatives}} \cr
& \frac{{dy}}{{dx}} = \frac{{{{\sec }^2}x}}{{\sqrt {{{\tan }^2}x + 1} }} \cr
& {\text{Using the identity ta}}{{\text{n}}^2}x + 1 = {\sec ^2}x \cr
& \frac{{dy}}{{dx}} = \frac{{{{\sec }^2}x}}{{\sqrt {{{\sec }^2}x} }} \cr
& {\text{Simplifying}} \cr
& \frac{{dy}}{{dx}} = \frac{{{{\sec }^2}x}}{{\sec x}} \cr
& \frac{{dy}}{{dx}} = \sec x \cr} $$
