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 - Page 188: 17



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: $...=e^{p}+\dfrac{3}{2}p^{1/2}e^{p}+pe^{p}+p^{3/2}e^{p}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.