Answer
$$f'\left( u \right) = \sec u$$
Work Step by Step
$$\eqalign{
& f\left( u \right) = {\sinh ^{ - 1}}\left( {\tan u} \right) \cr
& {\text{differentiate with respect to }}u \cr
& f'\left( u \right) = \frac{d}{{du}}\left[ {{{\sinh }^{ - 1}}\left( {\tan u} \right)} \right] \cr
& {\text{using the chain rule}}{\text{, }} \cr
& {\text{recall that }}\frac{d}{{du}}\left[ {{{\sinh }^{ - 1}}z} \right] = \frac{1}{{\sqrt {{z^2} + 1} }}\frac{{dz}}{{du}}.{\text{ consider }}z = \tan u \cr
& f'\left( u \right) = \frac{1}{{\sqrt {{{\left( {\tan u} \right)}^2} + 1} }}\frac{d}{{du}}\left[ {\tan u} \right] \cr
& {\text{solving the derivative}} \cr
& f'\left( u \right) = \frac{1}{{\sqrt {{{\left( {\tan u} \right)}^2} + 1} }}\left( {{{\sec }^2}u} \right) \cr
& f'\left( u \right) = \frac{{{{\sec }^2}u}}{{\sqrt {{{\tan }^2}u + 1} }} \cr
& {\text{by the pythagorean identity }}{\tan ^2}x + 1 = {\sec ^2}x \cr
& f'\left( u \right) = \frac{{{{\sec }^2}u}}{{\sqrt {{{\sec }^2}u} }} \cr
& f'\left( u \right) = \frac{{{{\sec }^2}u}}{{\sec u}} \cr
& f'\left( u \right) = \sec u \cr} $$
