Answer
$$ y' =\frac{1}{|t+1|\sqrt{(t+1)^2-1}}$$
Work Step by Step
Since $ y=\sec^{-1}(t+1)$, then the derivative $ y'$ is given by
$$ y'=\frac{1}{|t+1|\sqrt{(t+1)^2-1}}(+1)'=\frac{1}{|t+1|\sqrt{(t+1)^2-1}}.$$
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 - 7.8 Inverse Trigonometric Functions - Exercises - Page 374: 36
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