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: 29

Answer

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

Work Step by Step

$y= a+b$ where, $a = x^{2} u ^{4 }$ and, $ b = x v^{-2} $ where, $u = sin x$ $v = cos x$ Now, $u' = cos x$ $v' = -sin x$ $a' = 2xu^{4} + 4x^{2}u^{3} \times cos x = 2xsin^{4}x + 4x^{2}sin^{3}x cosx = 2xsin^{3}x(sin x + 2xcos x)$ $b' = v^{-2} - 2xv^{-3}(-sin x) = cos^{-2} x + 2xcos^{-3} x \times sin x = cos^{-2}x(1+2xcos^{-1} x \times sin x)$ So now, $\frac{dy}{dx} = a' +b'$ $\frac{dy}{dx} =2xsin^{3}x(sin x + 2xcos x) +cos^{-2}x(1+2xcos^{-1} x \times sin x) $
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