Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.5 The Chain Rule - 2.5 Exercises: 43

Answer

$$g'(x)=2r^2p\cos rx(2r\sin rx+n)^{p-1}$$

Work Step by Step

$$g'(x)=((2r\sin rx +n)^p)'=p(2r\sin rx+n)^{p-1}(2r\sin rx+n)'= p(2r\sin rx+n)^{p-1}(2r\cos rx \cdot(rx)'+0)= p(2r\sin rx+n)^{p-1}(2r\cos rx\cdot r)=2r^2p\cos rx(2r\sin rx+n)^{p-1}$$
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