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: 22

Answer

$\frac{dy}{dx} = 2e^{(4 \sqrt x + x^{2})} \times (\frac{1}{\sqrt x} + x)$

Work Step by Step

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