Answer
$$\eqalign{
& h'\left( t \right) = \left( { - 3\ln 0.02} \right)\left( {{{0.02}^t}} \right) \cr
& h''\left( t \right) = - 3\left( {{{0.02}^t}} \right){\left( {\ln 0.02} \right)^2} \cr} $$
Work Step by Step
$$\eqalign{
& h\left( t \right) = 7 - 3\left( {{{0.02}^t}} \right) \cr
& \cr
& {\text{Differentiate}} \cr
& h'\left( t \right) = \frac{d}{{dt}}\left( 7 \right) - 3\frac{d}{{dt}}\left( {{{0.02}^t}} \right) \cr
& {\text{compute the derivatives by the formula }}\left( {{a^t}} \right)' = {a^t}\ln a \cr
& h'\left( t \right) = - 3\left( {{{0.02}^t}} \right)\left( {\ln 0.02} \right) \cr
& h'\left( t \right) = \left( { - 3\ln 0.02} \right)\left( {{{0.02}^t}} \right) \cr
& \cr
& {\text{Differentiate to find the second derivative}} \cr
& h''\left( t \right) = \frac{d}{{dt}}\left( {\left( { - 3\ln 0.02} \right)\left( {{{0.02}^t}} \right)} \right) \cr
& h''\left( t \right) = \left( { - 3\ln 0.02} \right)\frac{d}{{dt}}\left( {{{0.02}^t}} \right) \cr
& h''\left( t \right) = \left( { - 3\ln 0.02} \right)\left( {{{0.02}^t}} \right)\left( {\ln 0.02} \right) \cr
& h''\left( t \right) = - 3\left( {{{0.02}^t}} \right){\left( {\ln 0.02} \right)^2} \cr} $$
