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: 69



Work Step by Step

$$\frac{dy}{dt}=\frac{d}{dt}\Big(3t(2t^2-5)^4\Big)$$ Following the Product Rule and Chain Rule: $$\frac{dy}{dt}=(2t^2-5)^4\frac{d}{dt}(3t)+(3t)\frac{d}{dt}(2t^2-5)^4$$ $$\frac{dy}{dt}=3(2t^2-5)^4+3t\times\Big(4(2t^2-5)^3\frac{d}{dt}(2t^2-5)\Big)$$ $$\frac{dy}{dt}=3(2t^2-5)^4+3t\times\Big(4(2t^2-5)^3(4t)\Big)$$ $$\frac{dy}{dt}=3(2t^2-5)^4+48t^2(2t^2-5)^3$$ $$\frac{dy}{dt}=(2t^2-5)^3\Big(3(2t^2-5)+48t^2\Big)$$ $$\frac{dy}{dt}=(2t^2-5)^3\Big(54t^2-15\Big)$$ $$\frac{dy}{dt}=3(2t^2-5)^3\Big(18t^2-5\Big)$$
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.