Answer
$y'=-\cot(x)$
Work Step by Step
$y=ln|csc(x)|$
$\frac{d}{dx}csc(x)=-(csc(x)cot(x))1$
$\frac{d}{dx}ln|x|=\frac{1}{x}$
$\frac{d}{dx}ln|csc(x)|=-\frac{csc(x)cot(x)}{csc(x)}$
$=\boxed{-\cot(x)}$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.1 Exercises - Page 325: 62
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