Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.6 - The Chain Rule - Exercises 3.6 - Page 149: 66



Work Step by Step

y=4$ sin(\sqrt{1+\sqrt{t}})$ Apply the chain rule: $\frac{dy}{dt}=4cos(\sqrt{1+\sqrt{t}}).\frac{d}{dt}(\sqrt{1+\sqrt{t}})$ =$4cos(\sqrt{1+\sqrt{t}}).\frac{1}{2\sqrt{\sqrt{1+\sqrt{t}}}}.\frac{d}{dt}(\sqrt{1+\sqrt{t}})$ =$\frac{cos(\sqrt{1+\sqrt{t}})}{\sqrt{t+t\sqrt{t}}}$
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