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 - Page 158: 12


$\frac{dy}{dx} = -\frac{5}{2} \frac{(\frac{\sqrt x}{2}-1)^{-11}}{\sqrt x}$

Work Step by Step

$y = u^{-10}$ where, $u = (\frac{\sqrt x}{2}-1)$ Now, $\frac{dy}{du} = -10u^{-11}$ and, $\frac{du}{dx} = \frac{1}{4 \sqrt x}$ So, $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$ $\frac{dy}{dx} = -10u^{-11} \times \frac{1}{4 \sqrt x}$ $\frac{dy}{dx} = -\frac{5}{2} \frac{(\frac{\sqrt x}{2}-1)^{-11}}{\sqrt x}$
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