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 - Page 211: 16

Answer

\[ = \frac{{{e^x}}}{x} + {e^x}\ln x\]

Work Step by Step

\[\begin{gathered} \frac{d}{{dx}}\,\left( {{e^x}\ln x} \right) \hfill \\ \hfill \\ Use\,\,product\,\,rule \hfill \\ \hfill \\ = {e^x}\frac{d}{{dx}}\,\,\left[ {\ln x} \right] + \ln x\frac{d}{{dx}}\,\left( {{e^x}} \right) \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ = {e^x}\,\left( {\frac{1}{x}} \right) + \ln x\,\left( {{e^x}} \right) \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = \frac{{{e^x}}}{x} + {e^x}\ln x \hfill \\ \end{gathered} \]
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