Calculus: Early Transcendentals (2nd Edition)

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

Chapter 3 - Derivatives - 3.3 Rules of Differentiation - 3.3 Exercises: 16


$g'(t) =\frac{3}{\sqrt t}$

Work Step by Step

$g(t) = 6\sqrt t$ $g'(t) = 8\times (\sqrt t)'$ $(\sqrt t)'=(t^{0.5})'= 0.5t^{-0.5}=\frac{1}{2t^{0.5}}=\frac{1}{2\sqrt t}$ Therefore $g'(t) = 6 \times (\sqrt t)'= 6\times \frac{1}{2\sqrt t}= \frac{3}{\sqrt t}$
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