Calculus: Early Transcendentals (2nd Edition)

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

Chapter 3 - Derivatives - 3.4 The Product and Quotient Rules - 3.4 Exercises: 14

Answer

$4e^t\left(\sqrt{t}+\frac{1}{2\sqrt{t}}\right)$

Work Step by Step

We are trying to find $\frac{d}{dt}4e^t\sqrt{t}$. Use the Product Rule, $\frac{d}{dt}u(t)v(t)=u'(t)v(t)+u(t)v'(t)$. In this case, $u(t)=4e^t$, so $u'(t)=4e^t$. Also, $v(t)=\sqrt{t}$, so $v'(w)=\frac{1}{2\sqrt{t}}$. Plugging these in, we get: $\frac{d}{dt}4e^t\sqrt{t}$ $=4e^t\sqrt{t}+4e^t*\frac{1}{2\sqrt{t}}$ $4e^t\left(\sqrt{t}+\frac{1}{2\sqrt{t}}\right)$
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