Answer
$$f'\left( t \right) = \frac{1}{{{{\tan }^{ - 1}}t\left( {1 + {t^2}} \right)}}$$
Work Step by Step
$$\eqalign{
& f\left( t \right) = \ln \left( {{{\tan }^{ - 1}}t} \right) \cr
& {\text{find the derivative}} \cr
& f'\left( t \right) = \frac{d}{{dt}}\left[ {\ln \left( {{{\tan }^{ - 1}}t} \right)} \right] \cr
& {\text{use }}\frac{d}{{dt}}\left[ {\ln u} \right] = \frac{1}{u}\frac{{du}}{{dt}}.{\text{ let }}u = {\tan ^{ - 1}}t \cr
& f'\left( t \right) = \frac{1}{{{{\tan }^{ - 1}}t}}\frac{d}{{dt}}\left[ {{{\tan }^{ - 1}}t} \right] \cr
& {\text{use }}\frac{d}{{dt}}\left[ {{{\tan }^{ - 1}}t} \right] = \frac{1}{{1 + {t^2}}} \cr
& f'\left( t \right) = \frac{1}{{{{\tan }^{ - 1}}t}}\left( {\frac{1}{{1 + {t^2}}}} \right) \cr
& {\text{simplifying}} \cr
& f'\left( t \right) = \frac{1}{{{{\tan }^{ - 1}}t\left( {1 + {t^2}} \right)}} \cr} $$
