Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - Chapter Review Exercises - Page 163: 47


$$ y'= \frac{(1+\sec t)-t(\sec t\tan t)}{(1+\sec t)^2}.$$

Work Step by Step

Recall the quotient rule: $(\dfrac{u}{v})'=\dfrac{vu'-uv'}{v^2}$ Recall that $(\sec x)'=\sec x\tan x$. Since $ y=\frac{t}{1+\sec t}$, then by the quotient rule, the derivative $ y'$ is given by $$ y'= \frac{(1+\sec t)-t(\sec t\tan t)}{(1+\sec t)^2}.$$
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.