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: 40

Answer

$$k'(x)=\sec\frac{1}{x}\Big(2x-\tan\frac{1}{x}\Big)$$

Work Step by Step

$$k(x)=x^2\sec\Big(\frac{1}{x}\Big)$$ The derivative of function $k(x)$ is: $$k'(x)=(x^2)'\sec\Big(\frac{1}{x}\Big)+x^2\Bigg[\sec\Big(\frac{1}{x}\Big)\Bigg]'$$ $$k'(x)=2x\sec\frac{1}{x}+x^2\sec\frac{1}{x}\tan\frac{1}{x}\Big(\frac{1}{x}\Big)'$$ $$k'(x)=2x\sec\frac{1}{x}+x^2\sec\frac{1}{x}\tan\frac{1}{x}\Big(\frac{-1(x)'}{x^2}\Big)$$ $$k'(x)=2x\sec\frac{1}{x}+x^2\sec\frac{1}{x}\tan\frac{1}{x}\Big(-\frac{1}{x^2}\Big)$$ $$k'(x)=2x\sec\frac{1}{x}-\sec\frac{1}{x}\tan\frac{1}{x}$$ $$k'(x)=\sec\frac{1}{x}\Big(2x-\tan\frac{1}{x}\Big)$$
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