Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 4 - Calculating the Derivative - 4.1 Techniques for Finding Derivatives - 4.1 Exercises - Page 207: 33

Answer

$$m = \frac{{25}}{6}$$

Work Step by Step

$$\eqalign{ & y = - 2{x^{1/2}} + {x^{3/2}},\,\,\,\,\,\,\,x = 9 \cr & {\text{find the derivative of the function}} \cr & \frac{{dy}}{{dx}} = {D_x}\left( { - 2{x^{1/2}} + {x^{3/2}}} \right) \cr & {\text{use sum rule for derivatives}} \cr & \frac{{dy}}{{dx}} = {D_x}\left( { - 2{x^{1/2}}} \right) + {D_x}\left( {{x^{3/2}}} \right) \cr & {\text{use the power rule for derivatives}} \cr & \frac{{dy}}{{dx}} = - 2\left( {\frac{1}{2}} \right){x^{1/2 - 1}} + \frac{3}{2}{x^{3/2 - 1}} \cr & \frac{{dy}}{{dx}} = - {x^{ - 1/2}} + \frac{3}{2}{x^{1/2}} \cr & \frac{{dy}}{{dx}} = - \frac{1}{{\sqrt x }} + \frac{3}{2}\sqrt x \cr & {\text{find the slope at }}x = 9{\text{ evaluating the derivative at }}x = 9 \cr & m = {\left. {\frac{{dy}}{{dx}}} \right|_{x = 9}} = - \frac{1}{{\sqrt 9 }} + \frac{3}{2}\sqrt 9 \cr & {\text{simplifying}} \cr & m = - \frac{1}{3} + \frac{9}{2} \cr & m = \frac{{ - 2 + 27}}{6} \cr & m = \frac{{25}}{6} \cr} $$
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