Answer
$h'\left( t \right) = 0$
Work Step by Step
$$\eqalign{
& h\left( t \right) = {\cot ^{ - 1}}t + {\cot ^{ - 1}}\left( {\frac{1}{t}} \right) \cr
& {\text{Differentiate}} \cr
& h'\left( t \right) = \frac{d}{{dt}}\left[ {{{\cot }^{ - 1}}t} \right] + \frac{d}{{dt}}\left[ {{{\cot }^{ - 1}}\left( {\frac{1}{t}} \right)} \right] \cr
& {\text{Apply }}\frac{d}{{dt}}\left[ {{{\cot }^{ - 1}}u} \right] = - \frac{1}{{1 + {u^2}}}\frac{{du}}{{dt}},{\text{ then}} \cr
& h'\left( t \right) = - \frac{1}{{1 + {t^2}}}\frac{d}{{dt}}\left[ t \right] - \frac{1}{{1 + {{\left( {1/t} \right)}^2}}}\frac{d}{{dt}}\left[ {\frac{1}{t}} \right] \cr
& h'\left( t \right) = - \frac{1}{{1 + {t^2}}}\left( 1 \right) - \frac{1}{{1 + {{\left( {1/t} \right)}^2}}}\left( { - \frac{1}{{{t^2}}}} \right) \cr
& {\text{Simplify}} \cr
& h'\left( t \right) = - \frac{1}{{1 + {t^2}}} + \frac{{{t^2}}}{{{t^2} + 1}}\left( {\frac{1}{{{t^2}}}} \right) \cr
& h'\left( t \right) = - \frac{1}{{1 + {t^2}}} + \frac{1}{{{t^2} + 1}} \cr
& h'\left( t \right) = 0 \cr} $$
