Answer
$$\eqalign{
& L'\left( t \right) = - \frac{{199.2{e^{ - 0.02t}}}}{{{{\left( {1 + 99.6{e^{ - 0.02t}}} \right)}^2}}} \cr
& L''\left( t \right) = \left( {\frac{{3.984 + 88.8384{e^{ - 0.02t}}}}{{{{\left( {1 + 99.6{e^{ - 0.02t}}} \right)}^3}}}} \right){e^{ - 0.02t}} \cr} $$
Work Step by Step
$$\eqalign{
& L\left( t \right) = \frac{{100}}{{1 + 99.6{e^{ - 0.02t}}}} \cr
& {\text{Differentiate}} \cr
& L'\left( t \right) = \frac{d}{{dt}}\left( {\frac{{100}}{{1 + 99.6{e^{ - 0.02t}}}}} \right) \cr
& {\text{use the quotient rule}} \cr
& L'\left( t \right) = \frac{{\left( {1 + 99.6{e^{ - 0.02t}}} \right)\left( {100} \right)' - 100\left( {1 + 99.6{e^{ - 0.02t}}} \right)'}}{{{{\left( {1 + 99.6{e^{ - 0.02t}}} \right)}^2}}} \cr
& L'\left( t \right) = \frac{{\left( {1 + 99.6{e^{ - 0.02t}}} \right)\left( 0 \right) - 100\left( {99.6{e^{ - 0.02t}}} \right)\left( { - 0.02t} \right)}}{{{{\left( {1 + 99.6{e^{ - 0.02t}}} \right)}^2}}} \cr
& L'\left( t \right) = - \frac{{199.2{e^{ - 0.02t}}}}{{{{\left( {1 + 99.6{e^{ - 0.02t}}} \right)}^2}}} \cr
& \cr
& {\text{Differentiate to find the second derivative}} \cr
& L''\left( t \right) = - \frac{d}{{dt}}\left( {\frac{{199.2{e^{ - 0.02t}}}}{{{{\left( {1 + 99.6{e^{ - 0.02t}}} \right)}^2}}}} \right) \cr
& {\text{use the quotient rule}} \cr
& L''\left( t \right) = - \frac{{{{\left( {1 + 99.6{e^{ - 0.02t}}} \right)}^2}\left( {199.2{e^{ - 0.02t}}} \right)' - \left( {199.2{e^{ - 0.02t}}} \right)\left( {{{\left( {1 + 99.6{e^{ - 0.02t}}} \right)}^2}} \right)'}}{{{{\left( {1 + 99.6{e^{ - 0.02t}}} \right)}^4}}} \cr
& = - \frac{{{{\left( {1 + 99.6{e^{ - 0.02t}}} \right)}^2}\left( { - 3.984{e^{ - 0.02t}}} \right) - \left( {199.2{e^{ - 0.02t}}} \right)\left( {2\left( {1 + 99.6{e^{ - 0.02t}}} \right)} \right)\left( {{e^{ - 0.02t}}} \right)\left( { - 0.02} \right)}}{{{{\left( {1 + 99.6{e^{ - 0.02t}}} \right)}^4}}} \cr
& {\text{simplifying}} \cr
& = - \frac{{\left( {1 + 99.6{e^{ - 0.02t}}} \right)\left( { - 3.984{e^{ - 0.02t}}} \right) - \left( {199.2{e^{ - 0.02t}}} \right)\left( 2 \right)\left( {{e^{ - 0.02t}}} \right)\left( { - 0.02} \right)}}{{{{\left( {1 + 99.6{e^{ - 0.02t}}} \right)}^3}}} \cr
& L''\left( t \right) = - \left( {\frac{{ - 3.984 - 396.8064{e^{ - 0.02t}} + 7.968{e^{ - 0.02t}}}}{{{{\left( {1 + 99.6{e^{ - 0.02t}}} \right)}^3}}}} \right){e^{ - 0.02t}} \cr
& L''\left( t \right) = - \left( {\frac{{ - 3.984 - 88.8384{e^{ - 0.02t}}}}{{{{\left( {1 + 99.6{e^{ - 0.02t}}} \right)}^3}}}} \right){e^{ - 0.02t}} \cr
& L''\left( t \right) = \left( {\frac{{3.984 + 88.8384{e^{ - 0.02t}}}}{{{{\left( {1 + 99.6{e^{ - 0.02t}}} \right)}^3}}}} \right){e^{ - 0.02t}} \cr} $$
