Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.9 - Linearization and Differentials - Exercises 3.9 - Page 175: 32

Answer

$\frac{-1}{2t\sqrt{t-1}}$

Work Step by Step

We have: $y=cot^{-1} \sqrt{t-1}=cot^{-1}(t-1)^{1/2}$ Derive: $\frac{dy}{dt}=\frac{(\frac{1}{2}(t-1)^{-1/2})}{1+[(t-1)^{1/2}]^{2}}$ =$\frac{-1}{2\sqrt{t-1}(1+t-1)}$ =$\frac{-1}{2t\sqrt{t-1}}$
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