## Calculus 8th Edition

Published by Cengage

# Chapter 6 - Inverse Functions - 6.3* The Natural Exponential Function - 6.3* Exercises - Page 453: 34

#### Answer

$e^r+er^{e-1}$

#### Work Step by Step

$\frac{d}{dr}(e^r+r^e)$ $=\frac{d}{dr}e^r+\frac{d}{dr}r^e$ $=e^r+er^{e-1}$

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