University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.6 - The Chain Rule - Exercises - Page 158: 23


$\frac{dp}{dt} = \frac{-1}{2\sqrt {3-t}}$

Work Step by Step

$p = \sqrt u$ where, $u = (3-t)$ Now, $\frac{dp}{du} =\frac{1}{2\sqrt u} $ and, $\frac{du}{dt} =-1$ So, $\frac{dp}{dt} = \frac{dp}{du} \times \frac{du}{dt}$ $\frac{dp}{dt} = \frac{-1}{2\sqrt u}$ $\frac{dp}{dt} = \frac{-1}{2\sqrt (3-t)}$
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