Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.5 Higher Derivatives - Exercises: 6

Answer

$y' = \dfrac{1}{2\sqrt{x}}$ $y'' = -\dfrac{1}{4\sqrt {x^3}}$ $y''' = \dfrac{3}{8\sqrt {x^5}}$

Work Step by Step

$y' = \dfrac{1}{2\sqrt{x}}$ $y'' = -\dfrac{1}{4\sqrt {x^3}}$ $y''' = \dfrac{3}{8\sqrt {x^5}}$
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