Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Practice Exercises - Page 177: 10



Work Step by Step

Take the derivative of the equation using Quotient Rule. Then apply Chain Rule to the inner terms: $s'=\frac{(t^{1/2}-1)(0)-(1)(\frac{1}{2}t^{-1/2}-0)}{(t^{1/2}-1)^2}$ $=\frac{-\frac{1}{2}t^{-1/2}}{(t^{1/2}-1)^2}$ $=\frac{-1}{2t^{1/2}(t^{1/2}-1)^2}$
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.