Answer
$e^r+er^{e-1}$
Work Step by Step
$\frac{d}{dr}e^r+r^e$
$=\frac{d}{dr}e^r+\frac{d}{dr}r^e$
$=e^r+er^{e-1}$
Calculus 8th Edition
by Stewart, James
Published by
Cengage
ISBN 10:
1285740629
ISBN 13:
978-1-28574-062-1
Chapter 6 - Inverse Functions - 6.2 Exponential Functions and Their Derivatives - 6.2 Exercises - Page 418: 32
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