Answer
$3\cos{x}-3\cos{x}+\sin{x}-\sin{x}=0$
Work Step by Step
$y'=\dfrac{d}{dx}(3\cos{x}+\sin{x})=-3\sin{x}+\cos{x}.$
$y''=\dfrac{d}{dx}(-3\sin{x}+\cos{x})--3\cos{x}-\sin{x}.$
$-3\cos{x}-\sin{x}+3\cos{x}+\sin{x}=0$
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 128: 128
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