Answer
$$ y'= \frac{1+e^x}{x+e^x}.$$
Work Step by Step
Recall that $(\ln x)'=\dfrac{1}{x}$
Recall that $(e^x)'=e^x$
Since $ y=\ln(x+e^x)$, then the derivative $ y'$ is given by
$$ y'=\frac{1}{x+e^x}(x+e^x)'=\frac{1+e^x}{x+e^x}.$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 7 - Exponential Functions - Chapter Review Exercises - Page 386: 16
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