Answer
$$ y'=\frac{-\csc^2 x}{\cot x}.$$
Work Step by Step
Recall that $(\ln x)'=\dfrac{1}{x}$
Recall that $(\cot x)'=-\csc^2 x$.
Since $ y=\ln \cot x $, then by the chain rule, we have
$$ y'=\frac{-\csc^2 x}{\cot x}.$$
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 - 7.3 Logarithms and Their Derivatives - Exercises - Page 343: 40
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