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: 18

Answer

$\frac{dy}{dx} = 20cos^{-5} x \times sin x $

Work Step by Step

$y = 5u^{-4}$ where, $u = cos x$ Now, $\frac{dy}{du} = -20u^{-5}$ and, $\frac{du}{dx} = -sin x$ So, $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$ $\frac{dy}{dx} = 20u^{-5} \times sin x$ $\frac{dy}{dx} = 20cos^{-5} x \times sin x $
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