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: 13



Work Step by Step

First, lets make an «u» substitution in order to make it easier. $f(u) = \sqrt{u}$ $u = 4 + \sqrt{3x}$ Then lets derivate using the chain rule $f'(u) = \dfrac{1}{2\sqrt{u}} (u') $ Now let's find u' $u'=\dfrac{3}{2\sqrt{3x}}$ Now let's undo the substitution and simplify $f'(x) = (\dfrac{1}{2\sqrt{4+\sqrt{3x}}})(\dfrac{3}{2\sqrt{3x}})$ And you got the answer: $f'(x)=\dfrac{3}{4\sqrt{3x}\sqrt{4+\sqrt{3x}}}$
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