Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.3 Rules of Differentiation - 3.3 Exercises: 33

Answer

$y'=\dfrac{1}{2\sqrt{x}}$

Work Step by Step

Simplify the expression: $y=\dfrac{x-a}{\sqrt{x}-\sqrt{a}} \times \dfrac{\sqrt{x}+\sqrt{a}}{\sqrt{x}+\sqrt{a}}$ $y=\dfrac{(x-a)(\sqrt{x}+\sqrt{a})}{(x-a)}=\sqrt{x}+\sqrt{a}$ $y=x^{1/2}+a^{1/2}$ Apply power rule: $y'=\dfrac{1}{2\sqrt{x}}$
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