University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Practice Exercises - Page 202: 60



Work Step by Step

Given $y=(1+t^2)\cot^{-1}{2t}$ On differentiating both sides: $\frac{dy}{dt}=\frac{d((1+t^2)\cot^{-1}{2t})}{dt}$ on applying multiply rulre of derivative $\frac{dy}{dt}={(1+t^2)}\frac{d(cot^{-1}2{t})}{dt}+\cot^{-1}{2t}\times{2t})$ $\frac{dy}{dt}={(1+t^2)}(\frac{-2}{1+(2t)^2})+2t\cot^{-1}{2t}$
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.