Answer
$$ y'=\frac{18x}{9x^2-8}.$$
Work Step by Step
Recall that $(\ln x)'=\dfrac{1}{x}$
Since $ y=\ln (9x^2-8)$, then we have
$$ y'=\frac{18x}{9x^2-8} .$$
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: 33
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