Answer
$f''(x)=\dfrac{48}{(x-2)^4}.$
Work Step by Step
$f(x)=8(x-2)^{-2}.$
$f'(x)=(8)(-2)(1)(x-2)^{-2-1}=-\dfrac{16}{(x-2)^3}=-16(x-2)^{-3}.$
$f''(x)=(-16)(-3)(1)(x-2)^{-3-1}=\dfrac{48}{(x-2)^4}.$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 2 - Differentiation - 2.4 Exercises - Page 137: 88
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