Calculus: Early Transcendentals 8th Edition

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

Chapter 3 - Section 3.2 - The Product and Quotient Rules - 3.2 Exercises: 4

Answer

$g'(x)=e^{x}(x+2\sqrt{x}+1+\dfrac{1}{\sqrt{x}})$

Work Step by Step

$g(x)=(x+2\sqrt{x})e^{x}$ Write the function in the following form: $g(x)=(x+2x^{1/2})e^{x}$ Differentiate by applying the product rule: $g'(x)=(x+2x^{1/2})(e^{x})'+(e^{x})(x+2x^{1/2})'=...$ $...=(x+2x^{1/2})(e^{x})+(e^{x})(1+2(\dfrac{1}{2})x^{-1/2})=...$ $...=(x+2x^{1/2})(e^{x})+(e^{x})(1+x^{-1/2})$ Take out common factor $e^{x}$: $g'(x)=e^{x}(x+2x^{1/2}+1+x^{-1/2})$ Rewrite like this for a better looking answer: $g'(x)=e^{x}(x+2\sqrt{x}+1+\dfrac{1}{\sqrt{x}})$
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