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: 23

Answer

$$\frac{{dy}}{{dx}} = \frac{{{e^x}}}{{\left| x \right|\sqrt {{x^2} - 1} }} + {e^x}{\sec ^{ - 1}}x$$

Work Step by Step

$$\eqalign{ & y = {e^x}{\sec ^{ - 1}}x \cr & {\text{differentiate using the product rule}} \cr & \frac{{dy}}{{dx}} = {e^x}\left( {{{\sec }^{ - 1}}x} \right)' + {\sec ^{ - 1}}x\left( {{e^x}} \right)' \cr & {\text{compute derivatives}} \cr & \frac{{dy}}{{dx}} = {e^x}\left( {\frac{1}{{\left| x \right|\sqrt {{x^2} - 1} }}} \right) + {\sec ^{ - 1}}x\left( {{e^x}} \right) \cr & {\text{simplify}} \cr & \frac{{dy}}{{dx}} = \frac{{{e^x}}}{{\left| x \right|\sqrt {{x^2} - 1} }} + {e^x}{\sec ^{ - 1}}x \cr} $$
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