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: 20

Answer

$y'=\dfrac{5}{2}z^{3/2}+\dfrac{1}{2}z^{-1/2}e^{z}+z^{1/2}e^{z}$

Work Step by Step

$y=(z^{2}+e^{z})\sqrt{z}$ Rewrite the function like this: $y=(z^{2}+e^{z})z^{1/2}$ Differentiate using the product rule: $y'=(z^{2}+e^{z})(z^{1/2})'+(z^{1/2})(z^{2}+e^{z})'=...$ $...=(z^{2}+e^{z})(\dfrac{1}{2}z^{-1/2})+(z^{1/2})(2z+e^{z})=...$ Evaluate the products and simplify $...=\dfrac{1}{2}z^{3/2}+\dfrac{1}{2}z^{-1/2}e^{z}+2z^{3/2}+z^{1/2}e^{z}=...$ $...=\dfrac{5}{2}z^{3/2}+\dfrac{1}{2}z^{-1/2}e^{z}+z^{1/2}e^{z}$
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