Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 2 - The Derivative - 2.6 The Chain Rule - Exercises Set 2.6 - Page 158: 14


$f'(x) = \dfrac{1}{6\sqrt{x}\sqrt[3] {(12+\sqrt{x})^2}}$

Work Step by Step

First, lets make an «u» substitution in order to make it easier. $f(u) = \sqrt[3] u$ $u =12+\sqrt{x}$ Then lets derivate using the chain rule $f'(u)=\dfrac{1}{3\sqrt[3] {u^2}} \times u'$ Now let's find u' $u' = \dfrac{1}{2\sqrt{x}}$ Now let's undo the substitution and simplify $f'(x) = \dfrac{1}{3\sqrt[3] {(12+\sqrt{x})^2}} \times \dfrac{1}{2\sqrt{x}} $ And you got the answer: $f'(x) = \dfrac{1}{6\sqrt{x}\sqrt[3] {(12+\sqrt{x})^2}}$
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