University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.11 - Linearization and Differentials - Exercises - Page 199: 27



Work Step by Step

We evaluate the function: $y=4\tan(\frac{{x}^3}{3})$ on differentiating the above: $\frac{dy}{dx}=\frac{d(4\tan(\frac{{x}^3}{3}))}{dx}$ or $\frac{dy}{dx}=4\sec^2(\frac{{x}^3}{3})\frac{d\frac{{x}^3}{3}}{dx}$ or $\frac{dy}{dx}=4\frac{3x^2}{3}\sec^2(\frac{{x}^3}{3})$ or ${dy}={4{x^2}}\sec^2({\frac{{x}^3}{3}}){dx}$ The final answer is: ${dy}={4{x^2}}\sec^2({\frac{{x}^3}{3}}){dx}$
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