Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.7 The Chain Rule - 3.7 Exercises - Page 191: 22


$y' =\frac{x}{\sqrt {x^2+9}} $

Work Step by Step

$y = \sqrt {x^2+9}$ $y' = (\sqrt {x^2+9})'$ The inner function is $g(x) = x^2+9$ and the outer function is $f(u) = \sqrt u$ $f'(u) = \frac{1}{2\sqrt u}$ Therefore by THEOREM 3.14 Version 2 $y' = (\sqrt {10x+1})' = \frac{1}{2\sqrt {g(x)}}g'(x) = \frac{1}{2\sqrt {x^2+9}}(2x) =\frac{x}{\sqrt {x^2+9}} $
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