Answer
$g'(\alpha)=\frac{5}{3}\cos{\alpha}-2.$
Work Step by Step
$g'(\alpha)=\dfrac{d}{d\alpha}(\frac{5}{3}\sin{\alpha})-\dfrac{d}{d\alpha}(2\alpha)\rightarrow$
$g'(\alpha)=\frac{5}{3}\cos{\alpha}-2.$
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 - Review Exercises - Page 157: 20
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