Calculus with Applications (10th Edition)

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

Chapter 4 - Calculating the Derivative - Chapter Review - Review Exercises: 44

Answer

\[{q^,} = 8{e^{2p + 1}}\,{\left( {{e^{2p + 1}} - 2} \right)^3}\]

Work Step by Step

\[\begin{gathered} q = \,{\left( {{e^{2p + 1}} - 2} \right)^4} \hfill \\ Find\,\,the\,\,derivative\,\,of\,\,the\,\,\,function \hfill \\ q = \,\,{\left[ {{{\left( {{e^{2p + 1}} - 2} \right)}^4}} \right]^,} \hfill \\ Use\,\,the\,\,general\,\,\,power\,\,rule \hfill \\ {q^,} = 4{\left( {{e^{2p + 1}} - 2} \right)^3}{\left( {{e^{2p + 1}} - 2} \right)^,} \hfill \\ Use\,\,\frac{d}{{dx}}\,\,\left[ {{e^{g\,\left( x \right)}}} \right] = {e^{g\,\left( x \right)}}{g^,}\,\left( x \right) \hfill \\ Then \hfill \\ {q^,} = 4{\left( {{e^{2p + 1}} - 2} \right)^3}\,\left( {2{e^{2p + 1}}} \right) \hfill \\ Multiplying \hfill \\ {q^,} = 8{e^{2p + 1}}\,{\left( {{e^{2p + 1}} - 2} \right)^3} \hfill \\ \hfill \\ \hfill \\ \end{gathered} \]
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