University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.9 - Inverse Trigonometric Functions - Exercises - Page 181: 51



Work Step by Step

given that ${y}={\sec^{-1}{x}}$ so ${x}={\sec{y}}$ differentiate the abve terms with respect to y: ${\frac{dx}{dy}}={\frac{d{\sec{y}}}{dy}} ={{\sec{y}{\tan{y}}}} $ ${\frac{dy}{dx}}={\frac{1}{\sec{y}\tan{y}}}$(on taking inverse of above term) ${[\frac{dy}{dx}]^2}={\frac{1}{\sec^2{y}\tan^2{y}}}$ (on squaring) ${[\frac{dy}{dx}]^2}={\frac{1}{\sec^2{y}(\sec^2{y}-1)}}$ ${[\frac{dy}{dx}]^2}={\frac{1}{x^2(x^2-1)}}$ on taking square root of both sides: ${\frac{dy}{dx}}={\frac{1}{|x|{\sqrt{(x^2-1)}}}}$ thus the final answer is: ${\frac{dy}{dx}}={\frac{1}{|x|{\sqrt{(x^2-1)}}}}$
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