University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.6 - The Chain Rule - Exercises - Page 158: 47



Work Step by Step

$$q=\sin\Big(\frac{t}{\sqrt{t+1}}\Big)$$ The derivative of function $q$ is $$q'=\cos\Big(\frac{t}{\sqrt{t+1}}\Big)\Big(\frac{t}{\sqrt{t+1}}\Big)'$$ We have $$\Big(\frac{t}{\sqrt{t+1}}\Big)'=\frac{(t)'\sqrt{t+1}-t(\sqrt{t+1})'}{t+1}=\frac{\sqrt{t+1}-\frac{t(t+1)'}{2\sqrt{t+1}}}{t+1}$$ $$=\frac{\sqrt{t+1}-\frac{t}{2\sqrt{t+1}}}{t+1}=\frac{\frac{2(t+1)-t}{2\sqrt{t+1}}}{t+1}=\frac{t+2}{2(t+1)\sqrt{t+1}}$$ Therefore, $$q'=\cos\Big(\frac{t}{\sqrt{t+1}}\Big)\frac{t+2}{2(t+1)\sqrt{t+1}}$$
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.