Calculus: Early Transcendentals (2nd Edition)

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

Chapter 3 - Derivatives - 3.4 The Product and Quotient Rules - 3.4 Exercises: 70

Answer

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

Work Step by Step

$y=\dfrac{x-a}{\sqrt{x}-\sqrt{a}}$ Factor the numerator as a difference of squares: $y=\dfrac{x-a}{\sqrt{x}-\sqrt{a}}=\dfrac{(\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a})}{\sqrt{x}-\sqrt{a}}=...$ Simplify: $...=\sqrt{x}+\sqrt{a}=...$ Rewrite the roots using rational exponents: $...=x^{1/2}+a^{1/2}$ Evaluate the derivative: $y'=\dfrac{1}{2}x^{-1/2}+0=\dfrac{1}{2x^{1/2}}=...$ $...=\dfrac{1}{2\sqrt{x}}$
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