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: 24


$f'(x) = \dfrac{3-8\sin(4x)\cos(4x)}{2\sqrt{3x-\sin^2(4x)}}$

Work Step by Step

First, lets make an «u» substitution in order to make it easier. $f(u) = \sqrt{u}$ $u = 3x - \sin^2(4x)$ Then lets derivate using the chain rule $f'(u) = \dfrac{u'}{2\sqrt{u}}$ Now let's find u' $u' = 3 -2(\sin(4x))\times 4\cos(4x)$ $u' = 3 - 8\sin(4x)\cos(4x)$ Now let's undo the substitution and simplify and you got the answer: $f'(x) = \dfrac{3-8\sin(4x)\cos(4x)}{2\sqrt{3x-\sin^2(4x)}}$
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