Calculus (3rd Edition)

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

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 $$ y'=\frac{18x}{9x^2-8} .$$
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