Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 6 - Exponential, Logarithmic, And Inverse Trigonometric Functions - 6.7 Derivatives And Integrals Involving Inverse Trigonometric Functions - Exercises Set 6.7 - Page 469: 20

Answer

$$\frac{{dy}}{{dx}} = \frac{5}{{\left| x \right|\sqrt {{x^{10}} - 1} }}$$

Work Step by Step

$$\eqalign{ & y = {\sec ^{ - 1}}\left( {{x^5}} \right) \cr & {\text{differentiate using the formula }} \cr & \frac{d}{{dx}}\left[ {{{\sec }^{ - 1}}u} \right] = \frac{1}{{\left| u \right|\sqrt {{u^2} - 1} }}\frac{{du}}{{dx}}{\text{ }}\left( {{\text{see page 467}}} \right) \cr & u = {x^5} \cr & \cr & \frac{{dy}}{{dx}} = \frac{1}{{\left| {{x^5}} \right|\sqrt {{{\left( {{x^5}} \right)}^2} - 1} }}\frac{d}{{dx}}\left( {{x^5}} \right) \cr & \frac{{dy}}{{dx}} = \frac{1}{{\left| {{x^5}} \right|\sqrt {{{\left( {{x^5}} \right)}^2} - 1} }}\left( {5{x^4}} \right) \cr & {\text{simplifying}}{\text{, }} \cr & \frac{{dy}}{{dx}} = \frac{{5{x^4}}}{{\left| {{x^5}} \right|\sqrt {{x^{10}} - 1} }} \cr & \frac{{dy}}{{dx}} = \frac{{5{x^4}}}{{{x^4}\left| x \right|\sqrt {{x^{10}} - 1} }} \cr & \frac{{dy}}{{dx}} = \frac{5}{{\left| x \right|\sqrt {{x^{10}} - 1} }} \cr} $$
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