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: 16

Answer

$y'=-\dfrac{3t^{2}+4t}{(t^{3}+2t^{2}-1)^{2}}$

Work Step by Step

$y=\dfrac{1}{t^{3}+2t^{2}-1}$ Differentiate using the quotient rule: $y'=\dfrac{(t^{3}+2t^{2}-1)(1)'-(1)(t^{3}+2t^{2}-1)'}{(t^{3}+2t^{2}-1)^{2}}=...$ $...=\dfrac{(t^{3}+2t^{2}-1)(0)-(1)(3t^{2}+4t)}{(t^{3}+2t^{2}-1)^{2}}=...$ Evaluate the products and simplify: $...=\dfrac{-3t^{2}-4t}{(t^{3}+2t^{2}-1)^{2}}=-\dfrac{3t^{2}+4t}{(t^{3}+2t^{2}-1)^{2}}$
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.