Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.9 Derivatives of Logarithmic and Exponential Functions - 3.9 Exercises: 28

Answer

\[\frac{{dp}}{{dt}} = \frac{{40\ln 2 \cdot {2^{ - t}}}}{{\,{{\left( {1 + {2^{ - t}}} \right)}^2}}}\]

Work Step by Step

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