Answer
$$\eqalign{
& g'\left( t \right) = 37\left( {\ln 1.05} \right)\left( {{{1.05}^t}} \right) \cr
& g''\left( t \right) = 37\left( {{{1.05}^t}} \right){\left( {\ln 1.05} \right)^2} \cr} $$
Work Step by Step
$$\eqalign{
& g\left( t \right) = 37\left( {{{1.05}^t}} \right) \cr
& {\text{Differentiate}} \cr
& g'\left( t \right) = \frac{d}{{dt}}\left( {37\left( {{{1.05}^t}} \right)} \right) \cr
& g'\left( t \right) = 37\frac{d}{{dt}}\left( {{{1.05}^t}} \right) \cr
& {\text{compute the derivative by the formula }}\left( {{a^t}} \right)' = {a^t}\ln a \cr
& g'\left( t \right) = 37\left( {{{1.05}^t}\ln 1.05} \right) \cr
& g'\left( t \right) = 37\left( {\ln 1.05} \right)\left( {{{1.05}^t}} \right) \cr
& \cr
& {\text{Differentiate to find the second derivative}} \cr
& g''\left( t \right) = \frac{d}{{dt}}\left( {37\left( {\ln 1.05} \right)\left( {{{1.05}^t}} \right)} \right) \cr
& g''\left( t \right) = 37\left( {\ln 1.05} \right)\frac{d}{{dt}}\left( {{{1.05}^t}} \right) \cr
& g''\left( t \right) = 37\left( {\ln 1.05} \right)\left( {{{1.05}^t}} \right)\left( {\ln 1.05} \right) \cr
& g''\left( t \right) = 37\left( {{{1.05}^t}} \right){\left( {\ln 1.05} \right)^2} \cr} $$
