Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.3 Differentiation Formulas - 2.3 Exercises: 36

Answer

$y'=\dfrac{\dfrac{2+x}{2\sqrt{x}}+\sqrt{x}}{4+4x+x^2}$

Work Step by Step

Identify a and b for apply quotient rule $a=\sqrt(x)$ $a'=\dfrac{1}{2\sqrt{x}}$ $b=2+x$ $b'=1$ Substitute in the formula $(\dfrac{a}{b})'=\dfrac{a'b-ab'}{b^2}$ $y'=\dfrac{(\dfrac{1}{2\sqrt{x}})(2+x)+(\sqrt{x})(1)}{(2+x)^2}$ $y'=\dfrac{\dfrac{2+x}{2\sqrt{x}}+\sqrt{x}}{4+4x+x^2}$
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