Answer
$y^{\prime \prime}=\sec ^{2} x$
Work Step by Step
$y=\ln |\sec x|\\
y^{\prime}=\frac{1}{\sec x} \times \sec x \tan x\\
y^{\prime}=\tan x\\
y^{\prime \prime}=\sec ^{2} x$
Calculus: Early Transcendentals 9th Edition
by Stewart, James
Published by
Cengage Learning
ISBN 10:
1337613924
ISBN 13:
978-1-33761-392-7
Chapter 3 - Section 3.6 - Derivatives of Logarithmic Functions - 3.6 Exercises - Page 224: 31
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