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: 33



Work Step by Step

$y=\sqrt{(\frac{{x^2+x}}{{x}^2})}=\sqrt{(\frac{{x+1}}{{x}})}$ on applying differentiation: $\frac{dy}{dx}=\frac{1}{2\sqrt{(\frac{{x+1}}{{x}})}}\frac{d}{dx}{(\frac{{x+1}}{{x}})}$ $\frac{dy}{dx}=\frac{1}{2\sqrt{(\frac{{x+1}}{{x}})}}(\frac{-(1+{x})\frac{d}{dx}({x})+{x}\frac{d}{dx}{(1+{x})}}{({x})^2})$ $\frac{dy}{dx}=(-\frac{\sqrt{x}}{2\sqrt{x+1}})(\frac{1}{({x})^2})$
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