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: 36

Answer

$$y'=\sin\frac{1}{x}-\frac{1}{x}\cos\frac{1}{x}$$

Work Step by Step

$$y'=(x\sin\frac{1}{x})'=x'\sin\frac{1}{x}+x(\sin\frac{1}{x})'=1\cdot\sin\frac{1}{x}+x\cdot\cos\frac{1}{x}(\frac{1}{x})'= \sin\frac{1}{x}+x\cos\frac{1}{x}(-\frac{1}{x^2})=\sin\frac{1}{x}-\frac{1}{x}\cos\frac{1}{x}$$
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