Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - Review Exercises - Page 158: 67

Answer

$f'(x) = \frac{-8x}{(x^{2}+1)^{2}}$ $f'(-1) = 2$

Work Step by Step

1. Find the derivative $f(x) = \frac{4}{(x^{2}+1)}$ $f(x) = 4(x^{2}+1)^{-1}$ $f'(x) = -4(x^{2}+1)^{-2}(2x)$ $= \frac{4}{(x^{2}+1)^{2}}(2x)$ $f'(x) = \frac{-8x}{(x^{2}+1)^{2}}$ 2. Evaluate the derivative at point $(-1, 2)$ $f'(-1) = \frac{-8(-1)}{((-1)^{2}+1)^{2}}$ $= \frac{8}{(1+1)^{2}}$ $= \frac{8}{(2)^{2}}$ $= \frac{8}{4}$ $= 2$
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