Calculus (3rd Edition)

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


$$ y' =\frac{1}{|t+1|\sqrt{(t+1)^2-1}}$$

Work Step by Step

Since $ y=\sec^{-1}(t+1)$, then the derivative $ y'$ is given by $$ y'=\frac{1}{|t+1|\sqrt{(t+1)^2-1}}(+1)'=\frac{1}{|t+1|\sqrt{(t+1)^2-1}}.$$
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