Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.1 - Derivatives of Polynomials and Exponential Functions - 3.1 Exercises: 26

Answer

$$k'(r)=e^r+er^{e-1}$$

Work Step by Step

$$k(r)=e^r+r^e$$ So, $$k'(r)=\frac{d}{dr}(e^r)+\frac{d}{dr}(r^e)$$ We have $\frac{d}{dr}(e^r)=e^r$ and $\frac{d}{dr}(r^e)=er^{e-1}$ Therefore, $$k'(r)=e^r+er^{e-1}$$
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