University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.11 - Linearization and Differentials - Exercises - Page 199: 28



Work Step by Step

We evaluate the function: $y=\sec({{x}^2-1})$ on differentiating the above: $\frac{dy}{dx}=\frac{d\sec({{x}^2-1})}{dx}$ or $\frac{dy}{dx}=\sec({{x}^2-1})\tan({{x}^2-1})\frac{d(({x})^2-1)}{dx}$ or $\frac{dy}{dx}={2{x}}\sec({{x}^2-1})\tan({{x}^2-1}))$ or ${dy}={2{x}}\sec({{x}^2-1})\tan({{x}^2-1})){dx}$ The final answer is: ${dy}={2{x}}\sec({{x}^2-1})\tan({{x}^2-1})){dx}$
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