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: 17

Answer

$y'=e^{p}+\dfrac{3}{2}p^{1/2}e^{p}+pe^{p}+p^{3/2}e^{p}$

Work Step by Step

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