Answer
$G'\left( u \right) = \frac{1}{{\sqrt {1 + {u^2}} }}$
Work Step by Step
$$\eqalign{
& G\left( u \right) = {\cosh ^{ - 1}}\sqrt {1 + {u^2}} ,{\text{ }}u > 0 \cr
& {\text{Differentiate}} \cr
& G'\left( u \right) = \frac{d}{{du}}\left[ {{{\cosh }^{ - 1}}\sqrt {1 + {u^2}} } \right] \cr
& {\text{Use Derivatives of Inverse Hyperbolic Functions }} \cr
& \frac{d}{{du}}\left[ {{{\cosh }^{ - 1}}v} \right] = \frac{1}{{\sqrt {{v^2} - 1} }}\frac{{dv}}{{du}},{\text{ let }}v = \sqrt {1 + {u^2}} ,{\text{ so}} \cr
& G'\left( u \right) = \frac{1}{{\sqrt {{{\left( {\sqrt {1 + {u^2}} } \right)}^2} - 1} }}\frac{d}{{du}}\left[ {\sqrt {1 + {u^2}} } \right] \cr
& {\text{Compute the derivative and simplify}} \cr
& G'\left( u \right) = \frac{1}{{\sqrt {1 + {u^2} - 1} }}\left( {\frac{{2u}}{{2\sqrt {1 + {u^2}} }}} \right) \cr
& G'\left( u \right) = \frac{1}{{\sqrt {{u^2}} }}\left( {\frac{u}{{\sqrt {1 + {u^2}} }}} \right) \cr
& G'\left( u \right) = \frac{1}{{\left| u \right|}}\left( {\frac{u}{{\sqrt {1 + {u^2}} }}} \right) \cr
& {\text{Where }}u > 0 \cr
& G'\left( u \right) = \frac{1}{u}\left( {\frac{u}{{\sqrt {1 + {u^2}} }}} \right) \cr
& G'\left( u \right) = \frac{1}{{\sqrt {1 + {u^2}} }} \cr} $$
