Answer
\[\frac{{dp}}{{dt}} = \frac{{40\ln 2 \cdot {2^{ - t}}}}{{\,{{\left( {1 + {2^{ - t}}} \right)}^2}}}\]
Work Step by Step
\[\begin{gathered}
p = \frac{{40}}{{1 + {2^{ - t}}}} \hfill \\
\hfill \\
Use\,\,the\,\,quotient\,\,rule \hfill \\
\hfill \\
\frac{{dp}}{{dt}} = \frac{{\,\left( {1 + {2^{ - t}}} \right)\,{{\left( {40} \right)}^,} - \,\left( {40} \right)\,{{\left( {1 + {2^{ - t}}} \right)}^,}}}{{\,{{\left( {1 + {2^{ - t}}} \right)}^2}}} \hfill \\
\hfill \\
differentiate,{\text{ u}}se\,\,the\,formula\,\,\frac{d}{{dx}}\,\,\left[ {{a^u}} \right] = {a^u}\ln a \cdot {u^,} \hfill \\
\hfill \\
\frac{{dp}}{{dt}} = \frac{{0 - 40\,\left( {{2^{ - t}}\ln 2 \cdot \,\left( { - 1} \right)} \right)}}{{\,{{\left( {1 + {2^{ - t}}} \right)}^2}}} \hfill \\
\hfill \\
multiply\;\;and\,\,simplify \hfill \\
\hfill \\
\frac{{dp}}{{dt}} = \frac{{40\ln 2 \cdot {2^{ - t}}}}{{\,{{\left( {1 + {2^{ - t}}} \right)}^2}}} \hfill \\
\hfill \\
\end{gathered} \]
