Thomas' Calculus 13th Edition

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

Chapter 3: Derivatives - Section 3.6 - The Chain Rule - Exercises 3.6 - Page 149: 72

Answer

$\frac{1}{2x}(1-\sqrt{x})^{-3}(\frac{3}{2}-\frac{1}{2\sqrt{x}})$

Work Step by Step

$y=(1-\sqrt{x})^{-1}$ Calculate the first derivative: $y'=-(1-\sqrt{x})^{-2}(-\frac{1}{2}x^{-1/2})$ =$\frac{1}{2}(1-\sqrt{x})^{-2}x^{-1/2}$ Now calculate the second derivative, using the first derivative above: $y''=\frac{1}{2}[(1-\sqrt{x})^{-2}(\frac{-1}{2}x^{-3/2})+x^{-1/2}(-2)((1-\sqrt{x})^{-3})(-\frac{1}{2}x^{-1/2})]$ =$\frac{1}{2}[\frac{-1}{2}x^{-3/2}(1-\sqrt{x})^{-2}+x^{-1}(1-\sqrt{x})^{-3}]$ =$\frac{1}{2}x^{-1}(1-\sqrt{x})^{-3}[\frac{-1}{2}x^{-1/2}(1-\sqrt{x})+1]$ =$\frac{1}{2x}(1-\sqrt{x})^{-1}(\frac{-1}{2\sqrt{x}}+\frac{1}{2}+1)$ =$\frac{1}{2x}(1-\sqrt{x})^{-3}(\frac{3}{2}-\frac{1}{2\sqrt{x}})$
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