Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.2 - The Product and Quotient Rules - 3.2 Exercises: 41

Answer

$f''(1) = \frac{1}{4}$

Work Step by Step

$f'(x) = \frac{2x(1+x) - (x^2)(1)}{(1+x)^2}$ $f'(x) = \frac{2x+x^2 - x^2}{(1+x)^2}$ $f'(x) = \frac{2x+x^2}{(1+x)^2}$ $f''(x) = \frac{(1+x)^2(2+2x)-[(-2x+x^2)(2(1+x))}{((1+x)^2)^2}$ $f''(x) = \frac{(1+x)^2(2+2x)-2(2x+x^2)}{(1+x)^4}$ Cancel out one $(1+x)$: $f''(x) = \frac{(1+x)(2+2x)-2(2x+x^2)}{(1+x)^3}$ $f''(1) = \frac{(1+1)(2+2(1))-2(2(1)+(1)^2)}{(1+1)^3}$ $f''(1) = \frac{2(4)-2(2+1)}{(2)^3}$ $f''(1) = \frac{8-6}{8}$ $f''(1) = \frac{2}{8}$ $f''(1) = \frac{1}{4}$
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