Answer
$$L'\left( t \right) = - \frac{{131.04{e^{3.9t}}}}{{{{\left( {1 + 2.1{e^{3.9t}}} \right)}^2}}}{\text{ and }}L'\left( t \right) = - \left( {\frac{{511.056 - 3219.6522{e^{3.9t}}}}{{{{\left( {1 + 2.1{e^{3.9t}}} \right)}^3}}}} \right){e^{3.9t}}$$
Work Step by Step
$$\eqalign{
& L\left( t \right) = \frac{{16}}{{1 + 2.1{e^{3.9t}}}} \cr
& {\text{Differentiate}} \cr
& L'\left( t \right) = \frac{d}{{dt}}\left( {\frac{{16}}{{1 + 2.1{e^{3.9t}}}}} \right) \cr
& {\text{use the quotient rule}} \cr
& L'\left( t \right) = \frac{{\left( {1 + 2.1{e^{3.9t}}} \right)\left( {16} \right)' - 16\left( {1 + 2.1{e^{3.9t}}} \right)'}}{{{{\left( {1 + 2.1{e^{3.9t}}} \right)}^2}}} \cr
& L'\left( t \right) = \frac{{\left( {1 + 2.1{e^{3.9t}}} \right)\left( 0 \right) - 16\left( {2.1{e^{3.9t}}} \right)\left( {3.9} \right)}}{{{{\left( {1 + 2.1{e^{3.9t}}} \right)}^2}}} \cr
& L'\left( t \right) = - \frac{{131.04{e^{3.9t}}}}{{{{\left( {1 + 2.1{e^{3.9t}}} \right)}^2}}} \cr
& \cr
& {\text{Differentiate to find the second derivative}} \cr
& L'\left( t \right) = - \frac{d}{{dt}}\left( {\frac{{131.04{e^{3.9t}}}}{{{{\left( {1 + 2.1{e^{3.9t}}} \right)}^2}}}} \right) \cr
& {\text{use the quotient rule}} \cr
& L'\left( t \right) = - \frac{{{{\left( {1 + 2.1{e^{3.9t}}} \right)}^2}\left( {131.04{e^{3.9t}}} \right)' - \left( {131.04{e^{3.9t}}} \right)\left( {{{\left( {1 + 2.1{e^{3.9t}}} \right)}^2}} \right)'}}{{{{\left( {1 + 2.1{e^{3.9t}}} \right)}^4}}} \cr
& L'\left( t \right) = - \frac{{{{\left( {1 + 2.1{e^{3.9t}}} \right)}^2}\left( {511.056{e^{3.9t}}} \right) - \left( {131.04{e^{3.9t}}} \right)\left( {2\left( {1 + 2.1{e^{3.9t}}} \right)} \right)\left( {8.19{e^{3.9t}}} \right)}}{{{{\left( {1 + 2.1{e^{3.9t}}} \right)}^4}}} \cr
& {\text{simplifying}} \cr
& L'\left( t \right) = - \frac{{\left( {1 + 2.1{e^{3.9t}}} \right)\left( {511.056{e^{3.9t}}} \right) - \left( {131.04{e^{3.9t}}} \right)\left( 2 \right)\left( {8.19{e^{3.9t}}} \right)}}{{{{\left( {1 + 2.1{e^{3.9t}}} \right)}^3}}} \cr
& L'\left( t \right) = - \left( {\frac{{\left( {1 + 2.1{e^{3.9t}}} \right)\left( {511.056} \right) - \left( {131.04{e^{3.9t}}} \right)\left( 2 \right)\left( {8.19} \right)}}{{{{\left( {1 + 2.1{e^{3.9t}}} \right)}^3}}}} \right){e^{3.9t}} \cr
& L'\left( t \right) = - \left( {\frac{{511.056 + 1073.2176{e^{3.9t}} - 2146.4352{e^{3.9t}}}}{{{{\left( {1 + 2.1{e^{3.9t}}} \right)}^3}}}} \right){e^{3.9t}} \cr
& L'\left( t \right) = - \left( {\frac{{511.056 - 3219.6522{e^{3.9t}}}}{{{{\left( {1 + 2.1{e^{3.9t}}} \right)}^3}}}} \right){e^{3.9t}} \cr} $$
