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: 31

Answer

$y' = \frac{cos(2\sqrt x)}{\sqrt x}$

Work Step by Step

$y = sin (2 \sqrt x)$ $y' = (sin (2 \sqrt x))'$ The inner function is $g(x) = 2\sqrt x$ and the outer function is $f(u) = sin(u)$ $f'(u) = cos(u)$ $y' = (sin (2 \sqrt x))' = cos(g(x))g'(x) = cos(2\sqrt x)\times\frac{2}{2\sqrt x}= \frac{cos(2\sqrt x)}{\sqrt x}$
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