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 - Page 192: 47

Answer

\[y' = {e^x}\cos \,\left( {\sin \,\left( {{e^x}} \right)} \right) \cdot \cos \,\left( {{e^x}} \right)\]

Work Step by Step

\[\begin{gathered} y = \sin \,\left( {\sin \,\left( {{e^x}} \right)} \right) \hfill \\ \hfill \\ Chain\,\,rule \hfill \\ \hfill \\ f\,{\left( {g\,\left( t \right)} \right)^,} = {f^,}\,\left( {g\,\left( t \right)} \right) \cdot {g^,}\,\left( t \right) \hfill \\ \hfill \\ then \hfill \\ \hfill \\ {y^,} = {\sin ^,}\,\left( {\sin \,\left( {{e^x}} \right)} \right) \cdot \,{\left( {\sin \,\left( {{e^x}} \right)} \right)^,} \hfill \\ \hfill \\ y' = \cos \,\left( {\sin \,\left( {{e^x}} \right)} \right) \cdot \cos \,\left( {{e^x}} \right) \cdot \,{\left( {{e^x}} \right)^,} \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ y' = {e^x}\cos \,\left( {\sin \,\left( {{e^x}} \right)} \right) \cdot \cos \,\left( {{e^x}} \right) \hfill \\ \end{gathered} \]
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