Answer
$y^{\prime}=\frac{a \sec ^{2}(\ln (a x+b))}{a x+b}$
Work Step by Step
$y=\tan (\ln (a x+b))\\
y^{\prime}=\sec ^{2}(\ln (a x+b)) \times \frac{1}{a x+b} \times a\\
y^{\prime}=\frac{a \sec ^{2}(\ln (a x+b))}{a x+b}$
Calculus: Early Transcendentals 8th Edition
by Stewart, James
Published by
Cengage Learning
ISBN 10:
1285741552
ISBN 13:
978-1-28574-155-0
Chapter 3 - Section 3.6 - Derivatives of Logarithmic Functions - 3.6 Exercises - Page 223: 21
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