Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.8 Inverse Trigonometric Functions - Exercises - Page 374: 47


$$ y'= - \frac{ 1}{ x\sqrt{1-(\ln x)^2}}.$$

Work Step by Step

Since $ y=\ arcos( \ln x)=\cos^{-1}( \ln x) $, then $$ y'=- \frac{1}{ \sqrt{1-(\ln x)^2}}(\ln x)'\\ =- \frac{ 1/x}{ \sqrt{1-(\ln x)^2}}=- \frac{ 1}{ x\sqrt{1-(\ln x)^2}}.$$
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