Answer
$y'=1-\csc^2(x)$.
Work Step by Step
$y=g(x)+h(x)\rightarrow g(x)=x$ and $h(x)=\cot(x)$.
Using the Power Rule: $g'(x)=1$.
By Theorem $2.9$: $h'(x)=\frac{d}{dx}(\cot(x))=-\csc^2(x)$.
Using the Sum Rule $y'=g'(x)+h'(x)$.
$y'=1-\csc^2(x)$.
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 2 - Differentiation - 2.3 Exercises - Page 125: 44
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