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,
$$k'(r)=e^r+er^{e-1}$$
Calculus: Early Transcendentals 9th Edition
by Stewart, James
Published by
Cengage Learning
ISBN 10:
1337613924
ISBN 13:
978-1-33761-392-7
Chapter 3 - Section 3.1 - Derivatives of Polynomials and Exponential Functions - 3.1 Exercises - Page 182: 28
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