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: 33

Answer

$G'(x)=-\dfrac{C(\ln4)4^{C/x}}{x^{2}}$

Work Step by Step

$G(x)=4^{C/x}$ (Here, $C$ is a constant) Differentiate using the chain rule: $G'(x)=4^{C/x}(\ln4)(\dfrac{C}{x})'=...$ Rewrite $\dfrac{C}{x}$ as $Cx^{-1}$ and continue with the differentiation process: $...=4^{C/x}(\ln4)(\dfrac{C}{x})'=4^{C/x}(\ln4)(Cx^{-1})'=4^{C/x}(\ln4)(-Cx^{-2})$ $G'(x)=-\dfrac{C(\ln4)4^{C/x}}{x^{2}}$
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