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: 37

Answer

\[{g^,}\,\left( y \right) = {e^{y + 1}}{y^{e - 1}} + {e^y}{y^e}\]

Work Step by Step

\[\begin{gathered} g\,\left( y \right) = {e^y} \cdot {y^e} \hfill \\ \hfill \\ Use\,\,the\,\,product\,\,rule \hfill \\ \hfill \\ {g^,}\,\left( y \right) = {e^y}\,{\left( {{y^e}} \right)^,} + \,\left( {{y^e}} \right)\,{\left( {{e^y}} \right)^,} \hfill \\ \hfill \\ differentiate \hfill \\ \hfill \\ {g^,}\,\left( y \right) = {e^y}\,\,\left( {e{y^{e - 1}}} \right) + {y^e}\,\left( {{e^y}} \right) \hfill \\ \hfill \\ multiply \hfill \\ \hfill \\ {g^,}\,\left( y \right) = {e^{y + 1}}{y^{e - 1}} + {e^y}{y^e} \hfill \\ \end{gathered} \]
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