Calculus: Early Transcendentals (2nd Edition)

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

Chapter 3 - Derivatives - 3.7 The Chain Rule - 3.7 Exercises: 64

Answer

\[{y^,} = 2{e^{2x \cdot \,}} \cdot \,\,\,{\left( {2x - 7} \right)^5} + 10{e^{2x}}\, \cdot \,\,{\left( {2x - 7} \right)^4}\]

Work Step by Step

\[\begin{gathered} y = {e^{2x}}\,{\left( {2x - 7} \right)^5} \hfill \\ \hfill \\ Product\,\,rule \hfill \\ \hfill \\ {y^,} = \,{\left( {{e^{2x}}} \right)^,} \cdot \,{\left( {2x - 7} \right)^5} + {e^{2x}} \cdot \,{\left( {\,{{\left( {2x - 7} \right)}^5}} \right)^,} \hfill \\ \hfill \\ Chain\,\,rule \hfill \\ \hfill \\ {y^,} = {e^{2x}}\,{\left( {2x} \right)^,} \cdot \,\,{\left( {2x - 7} \right)^5} + {e^{2x}} \cdot 5\,{\left( {2x - 7} \right)^4} \cdot \,\,{\left( {2x - 7} \right)^,} \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ {y^,} = 2{e^{2x}} \cdot \,{\left( {2x - 7} \right)^5} + 5{e^{2x}} \cdot \,{\left( {2x - 7} \right)^4} \cdot \,\,\,2 \hfill \\ \hfill \\ multiply \hfill \\ \hfill \\ {y^,} = 2{e^{2x \cdot \,}} \cdot \,\,\,{\left( {2x - 7} \right)^5} + 10{e^{2x}}\, \cdot \,\,{\left( {2x - 7} \right)^4} \hfill \\ \end{gathered} \]
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