Answer
$$ y'=\frac{1}{x+1}- \frac{3x^2}{x^3+1}.$$
Work Step by Step
Since $ y=\ln\left( \frac{x+1}{x^3+1}\right)$, first we simplify $ y $ as follows
$$ y=\ln (x+1)- \ln (x^3+1)$$
then we have
$$ y'=\frac{1}{x+1}- \frac{3x^2}{x^3+1}.$$
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 - 7.3 Logarithms and Their Derivatives - Exercises - Page 343: 44
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