Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - Review - Exercises - Page 199: 87

Answer

$\frac{1}{32}$

Work Step by Step

The limit can be written: $$\begin{align*} \lim\limits_{h \to 0}\frac{\sqrt[4] {16+h}-2}{h}& =\lim\limits_{h \to 0}\frac{\sqrt[4] {16+h}-\sqrt[4]{16}}{h} \\ &=\left[\frac{d}{dx}\sqrt[4] x\right]_{x=16}\\ & = \frac{1}{4}x^{-\frac{3}{4}}|_{x=16}\\ & = \frac{1}{32} \end{align*}$$
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