Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 4 - Calculating the Derivative - 4.4 Derivatives of Exponential Functions - 4.4 Exercises: 22

Answer

\[{p^,} = \frac{{1250{e^{ - 0.5t}}}}{{\,{{\left( {12 + 5{e^{ - 0.5t}}} \right)}^2}}}\]

Work Step by Step

\[\begin{gathered} p = \frac{{500}}{{12 + 5{e^{ - 0.5t}}}} \hfill \\ Find\,\,the\,\,derivative \hfill \\ {p^,} = \frac{d}{{dt}}\,\,\left[ {\,\frac{{500}}{{12 + 5{e^{ - 0.5t}}}}} \right] \hfill \\ Use\,\,the\,\,quotient\,\,rule \hfill \\ {p^,} = \frac{{\,\left( {12 + 5{e^{ - 0.5t}}} \right)\,{{\left( {500} \right)}^,} - 500\,{{\left( {12 + 5{e^{ - 0.5t}}} \right)}^,}}}{{\,{{\left( {12 + 5{e^{ - 0.5t}}} \right)}^2}}} \hfill \\ Then \hfill \\ {p^,} = \frac{{\,\left( {12 + 5{e^{ - 0.5t}}} \right)\,\left( 0 \right) - 500\,\left( {5{e^{ - 0.5t}}} \right)\,\left( { - 0.5} \right)}}{{\,{{\left( {12 + 5{e^{ - 0.5t}}} \right)}^2}}} \hfill \\ Multiply \hfill \\ {p^,} = \frac{{1250{e^{ - 0.5t}}}}{{\,{{\left( {12 + 5{e^{ - 0.5t}}} \right)}^2}}} \hfill \\ \end{gathered} \]
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