University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Practice Exercises - Page 202: 23


$y'=-\dfrac{1}{2}x^{-3/2}\sec(2x)^2 +8x^{1/2}\sec(2x)^2\tan(2x)^2 $

Work Step by Step

Given that $y=(x)^{-1/2}\sec(2x)^2 $ Apply derivative rules of differentiation: $f(x)=p'(x)q(x)+p(x)q'(x)$ $y'=\dfrac{d}{dx}[(x)^{-1/2}\sec(2x)^2 ]$ $=-\dfrac{1}{2x^{3/2}}\sec(2x)^2 +\dfrac{1}{x^{1/2}}\sec(2x)^2\tan(2x)^2 ]\dfrac{d}{dx}[(2x)^2 ]$ $=-\dfrac{1}{2x^{3/2}}\sec(2x)^2 +\dfrac{1}{x^{1/2}}\sec(2x)^2\tan(2x)^2 ](8x)$ Hence, $y'=-\dfrac{1}{2}x^{-3/2}\sec(2x)^2 +8x^{1/2}\sec(2x)^2\tan(2x)^2 $
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