Answer
$$ y'= - \frac{ 1}{ x\sqrt{1-(\ln x)^2}}.$$
Work Step by Step
Since $ y=\ arcos( \ln x)=\cos^{-1}( \ln x) $, then
$$ y'=- \frac{1}{ \sqrt{1-(\ln x)^2}}(\ln x)'\\
=- \frac{ 1/x}{ \sqrt{1-(\ln x)^2}}=- \frac{ 1}{ x\sqrt{1-(\ln x)^2}}.$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
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
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