University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.3 - Differentiation Rules - Exercises: 46

Answer

$s' = \frac{-5t+2}{t^{3}}$ $s'' = \frac{2(5t-3)}{t^{4}}$

Work Step by Step

$s = \frac{t^{2}+5t-1}{t^{2}}$ $s' = \frac{t^{2}(2t+5) - 2t(t^{2}+5t-1)}{t^{4}}$ $s' = \frac{-5t^{2}+2t}{t^{4}}$ $s' = \frac{-5t+2}{t^{3}}$ $s'' = \frac{t^{3}(-5)-3t^{2}(-5t+2)}{t^{6}}$ $s'' = \frac{10t^{3}-6t^{2}}{t^{6}}$ $s'' = \frac{10t-6}{t^{4}}$ $s'' = \frac{2(5t-3)}{t^{4}}$
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.