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: 64


$f'(x)=\frac{2}{(x+2)^2}$ $f''(x)=\frac{-4}{(x+2)^3}$

Work Step by Step

$f(x)=\frac{x}{x+2}$ Using Quotient Rule, where $(\frac{f}{g})'=\frac{f'g-fg'}{g^2}$ $f'(x)=\frac{(1)(x+2)-(x)(1)}{(x+2)^2}=\frac{2}{(x+2)^2}$ Using Quotient Rule and Chain Rule: $f''(x)=\frac{(0)(x+2)^2-2(2)(1)(x+2)}{(x+2)^4}=\frac{-4}{(x+2)^3}$
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