Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.5 The Chain Rule - 2.5 Exercises: 38

Answer

$$y'=\cos(t+\cos\sqrt t)(1-\frac{1}{2\sqrt t}\sin\sqrt t)$$

Work Step by Step

$$y'=(\sin(t+\cos\sqrt t))'=\cos(t+\cos\sqrt t)(t+\cos\sqrt t)'= \cos(t+\cos \sqrt t)(1+(-\sin\sqrt t)(\sqrt t)')= \cos(t+\cos\sqrt t)(1-\frac{1}{2\sqrt t}\sin\sqrt t)$$
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