Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.4 - The Chain Rule - 3.4 Exercises: 27

Answer

$r'(t)=\dfrac{(\ln10)10^{2\sqrt{t}}}{\sqrt{t}}$

Work Step by Step

$r(t)=10^{2\sqrt{t}}$ Differentiate using the chain rule: $r'(t)=(10^{2\sqrt{t}})(\ln10)(2\sqrt{t})'$ Rewriting $2\sqrt{t}$ as $2t^{1/2}$, we proceed with the differentiation process: $r'(t)=(10^{2\sqrt{t}})(\ln10)(2\sqrt{t})'=(10^{2\sqrt{t}})(\ln10)(2t^{1/2})'=...$ $...=(10^{2\sqrt{t}})(\ln10)(2\dfrac{1}{2}t^{-1/2})=(10^{2\sqrt{t}})(\ln10)(t^{-1/2})$ Simplifying: $r'(t)=\dfrac{(\ln10)10^{2\sqrt{t}}}{\sqrt{t}}$
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