University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.2 - The Derivative as a Function - Exercises: 5

Answer

$\displaystyle p'(x) = \frac{3}{2\sqrt {3x}}$ $\displaystyle p'(1) = \frac{3}{2\sqrt{3}}=\frac{\sqrt 3}{2}$ $\displaystyle p'(3) = \frac{1}{2}$ $\displaystyle p'(\frac{2}{3}) = \frac{3}{2\sqrt 2}$

Work Step by Step

$p(x) =\sqrt (3x)$ Using the derivative definition: $p'(x) = \lim\limits_{h \to 0}\frac{\sqrt (3(x+h)) - \sqrt (3x)}{h}$ $p'(x) = \lim\limits_{h \to 0}\frac{3(x+h) - 3x}{h(\sqrt (3(x+h)) + \sqrt (3x))}$ $p'(x) = \frac{3}{2\sqrt (3x)}$ Now, $p'(1) = \frac{\sqrt 3}{2}$ $p'(3) = \frac{1}{2}$ $p'(\frac{2}{3}) = \frac{3}{2\sqrt 2}$
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