Calculus 10th Edition

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

Chapter 2 - Differentiation - 2.4 Exercises: 92

Answer

$f''(0)=\dfrac{3}{128}.$

Work Step by Step

$f(x)=(x+4)^{-\frac{1}{2}}$ $f'(x)=-\dfrac{1}{2}(x+4)^{-\frac{1}{2}-1}=-\dfrac{1}{2\sqrt{(x+4)^3}}$ $f'(x)=-\dfrac{1}{2}(x+4)^-\frac{3}{2}$ $f''(x)=(-\dfrac{1}{2})(-\dfrac{3}{2})(x+4)^{-\frac{3}{2}-1}=\dfrac{3}{4\sqrt{(x+4)^5}}$ $f''(0)=\dfrac{3}{4\sqrt{(0+4)^5}}=\dfrac{3}{128}.$
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