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: 30


$\frac{dy}{dx} = -\frac{1}{x} sin ^{-5}(\frac{1}{x} + 5sin ^{-1}x cos x) - cos^{2}x (\frac{1}{3}cos x -x sin x)$

Work Step by Step

$y = \frac{1}{x}u^{-5} - \frac{x}{3}v^{3}$ where, $u = sin x$ $v = cos x$ and, $u' = cos x$ $v' = -sin x$ Now, $y = a-b$, where, $a = \frac{1}{x}u^{-5}$ $b = \frac{x}{3}v^{3}$ Differentiate these two terms: $a' = \frac{-1}{x^{2}}u^{-5} -5 \frac{1}{x}u^{-6} \times (cos x)$ $ = \frac{-1}{x^{2}}sin^{-5} x -5 \frac{1}{x}sin^{-6} x cos x$ $ = -\frac{1}{x} sin ^{-5}(\frac{1}{x} + 5sin ^{-1}x cos x)$ $b' = \frac{1}{3}v^{3} + xv^{2} \times (-sin x) = \frac{1}{3}cos^{3}x - xcos^{2}x sin x$ $b' = cos^{2}x (\frac{1}{3}cos x -x sin x)$ So now, $\frac{dy}{dx} = a' - b'$ $\frac{dy}{dx} = -\frac{1}{x} sin ^{-5}(\frac{1}{x} + 5sin ^{-1}x cos x) - cos^{2}x (\frac{1}{3}cos x -x sin x)$
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