Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.4 - The Chain Rule - 3.4 Exercises - Page 204: 43

Answer

$g'(x)=2pr^{2}\ln a(2ra^{rx}+n)^{p-1}a^{rx}$

Work Step by Step

$g(x)=(2ra^{rx}+n)^p$ Differentiate using the chain rule: $g'(x)=p(2ra^{rx}+n)^{p-1}(2ra^{rx}+n)'=...$ $...=p(2ra^{rx}+n)^{p-1}(2ra^{rx})'=...$ $...=2pr(2ra^{rx}+n)^{p-1}(a^{rx})'=...$ Use the chain rule one more time to find $(a^{rx})'$: $...=2pr(2ra^{rx}+n)^{p-1}(a^{rx})(\ln a)(rx)'=...$ $...=2pr(2ra^{rx}+n)^{p-1}(a^{rx})(\ln a)(r)=...$ $...=2pr^{2}\ln a(2ra^{rx}+n)^{p-1}a^{rx}$
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