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: 30



Work Step by Step

We evaluate the function: $y=2\cot(\frac{1}{\sqrt{x}})$ $y=2\cot({{x}^{\frac{-1}{2}}})$ On differentiating the above: $\frac{dy}{dx}=\frac{d(2{\cot{x^\frac{-1}{2}}}))}{dx}$ or $\frac{dy}{dx}=-2\csc^2({{x}^{(\frac{-1}{2})}})\frac{d(({{{x}^{(\frac{-1}{2})}}})}{dx}$ or $\frac{dy}{dx}={{x}^{(\frac{-3}{2})}}\csc^2({{x}^{(\frac{-1}{2})}})$ or ${dy}={{x}^{(\frac{-3}{2})}}\csc^2({{x}^{(\frac{-1}{2})}}){dx}$ The final answer is: ${dy}={{x}^{(\frac{-3}{2})}}\csc^2({{x}^{(\frac{-1}{2})}}){dx}$
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