Answer
$F'\left( t \right) = \frac{{t{{\sec }^2}\sqrt {1 + {t^2}} }}{{\sqrt {1 + {t^2}} }}$
Work Step by Step
$$\eqalign{
& F\left( t \right) = \tan \sqrt {1 + {t^2}} \cr
& {\text{Differentiate}} \cr
& F'\left( t \right) = \frac{d}{{dt}}\left[ {\tan \sqrt {1 + {t^2}} } \right] \cr
& {\text{Apply the chain rule}}{\text{, recall that }}\frac{d}{{dx}}\left[ {\tan x} \right] = {\sec ^2}x,{\text{ then}} \cr
& F'\left( t \right) = {\sec ^2}\sqrt {1 + {t^2}} \frac{d}{{dt}}\left[ {\sqrt {1 + {t^2}} } \right] \cr
& {\text{Apply the chain rule }} \cr
& F'\left( t \right) = {\sec ^2}\sqrt {1 + {t^2}} \frac{d}{{dt}}\left[ {\sqrt {1 + {t^2}} } \right] \cr
& {\text{Apply the general power rule}} \cr
& F'\left( t \right) = {\sec ^2}\sqrt {1 + {t^2}} \left( {\frac{1}{2}} \right){\left( {1 + {t^2}} \right)^{ - 1/2}}\frac{d}{{dt}}\left[ {1 + {t^2}} \right] \cr
& {\text{Differentiate and simplify}} \cr
& F'\left( t \right) = {\sec ^2}\sqrt {1 + {t^2}} \left( {\frac{1}{2}} \right){\left( {1 + {t^2}} \right)^{ - 1/2}}\left( {2t} \right) \cr
& F'\left( t \right) = {\sec ^2}\sqrt {1 + {t^2}} {\left( {1 + {t^2}} \right)^{ - 1/2}}t \cr
& F'\left( t \right) = \frac{{t{{\sec }^2}\sqrt {1 + {t^2}} }}{{\sqrt {1 + {t^2}} }} \cr} $$
