Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 2 - The Derivative - 2.6 The Chain Rule - Exercises Set 2.6: 54

Answer

$$\frac{d^2y}{dx^2}=\frac{2}{x^3}\sec^2\frac{1}{x}\tan\frac{1}{x}$$

Work Step by Step

$$\frac{dy}{dx}=\Big(x\tan\frac{1}{x}\Big)'=x'\tan\frac{1}{x}+x\Big(\tan\frac{1}{x}\Big)'= \tan\frac{1}{x}+x\sec^2\frac{1}{x}\Big(\frac{1}{x}\Big)'=\tan\frac{1}{x}+x\sec^2\frac{1}{x}\Big(-\frac{1}{x^2}\Big)= \tan\frac{1}{x}-\frac{1}{x}\sec^2\frac{1}{x}$$ $$\frac{d^2y}{dx^2}=\frac{d}{dx}\Big(\frac{dy}{dx}\Big)=\Big(\tan\frac{1}{x}-\frac{1}{x}\sec^2\frac{1}{x}\Big)'= \sec^2\frac{1}{x}\Big(\frac{1}{x}\Big)'-\Big(\Big(\frac{1}{x}\Big)'\sec^2\frac{1}{x}+\frac{1}{x}\Big(\sec^2\frac{1}{x}\Big)'\Big)= \sec^2\frac{1}{x}\Big(-\frac{1}{x^2}\Big)-\Big(-\frac{1}{x^2}\sec^2\frac{1}{x}\Big)-\frac{1}{x}\cdot2\sec\frac{1}{x}\Big(\sec\frac{1}{x}\Big)'= -\frac{2}{x}\sec\frac{1}{x}\sec\frac{1}{x}\tan\frac{1}{x}\Big(\frac{1}{x}\Big)'= -\frac{2}{x}\sec^2\frac{1}{x}\tan\frac{1}{x}\Big(-\frac{1}{x^2}\Big)= \frac{2}{x^3}\sec^2\frac{1}{x}\tan\frac{1}{x}$$
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