University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Practice Exercises - Page 202: 61



Work Step by Step

Given $y=z\sec^{-1}{z}-\sqrt{(z^2-1)}$ On differentiating both sides: $\frac{dy}{dz}=\frac{d(z\sec^{-1}{z}-\sqrt{(z^2-1)})}{dz}$ $\frac{dy}{dz}={\frac{z}{|z|\sqrt{(z^2-1)}}}+sec^{-1}{z}-\frac{1}{2\sqrt{(z^2-1)}}\frac{d({(z^2-1)})}{dz}$ $\frac{dy}{dz}={\frac{z}{|z|\sqrt{(z^2-1)}}}+sec^{-1}{z}-\frac{z}{\sqrt{(z^2-1)}}$ $\frac{dy}{dz}={\frac{1-z}{\sqrt{(z^2-1)}}}+sec^{-1}{z}$
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