University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.6 - The Chain Rule - Exercises: 14

Answer

$\frac{dy}{dx} = \frac{3x-2}{\sqrt {3x^{2} - 4x +6}} $

Work Step by Step

$y = \sqrt u$ where, $u = 3x^{2} - 4x +6$ Now, $\frac{dy}{du} = \frac{1}{2 \sqrt u} $ and, $\frac{du}{dx} = 6x -4$ So, $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$ $\frac{dy}{dx} = \frac{1}{2 \sqrt u} \times (6x-4)$ $\frac{dy}{dx} = \frac{1}{\sqrt (3x^{2} - 4x +6)} \times (3x-2)$
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