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 - Page 161: 76

Answer

$$9$$

Work Step by Step

$$\eqalign{ & {\text{Calculate the derivative }}\frac{d}{{dx}}\left[ {xf\left( x \right)} \right]{\text{ using the product rule}} \cr & \frac{d}{{dx}}\left[ {xf\left( x \right)} \right] = \left( 1 \right)f\left( x \right) + xf'\left( x \right) \cr & \frac{d}{{dx}}\left[ {xf\left( x \right)} \right] = f\left( x \right) + xf'\left( x \right) \cr & {\text{Calculate the derivative at }}x = 3 \cr & {\left. {\frac{d}{{dx}}\left[ {xf\left( x \right)} \right]} \right|_{x = 3}} = f\left( 3 \right) + 3f'\left( 3 \right) \cr & {\text{From the table we know that }}f\left( 3 \right) = 3{\text{ and }}f'\left( 3 \right) = 2 \cr & {\left. {\frac{d}{{dx}}\left[ {xf\left( x \right)} \right]} \right|_{x = 3}} = 3 + 3\left( 2 \right) \cr & {\left. {\frac{d}{{dx}}\left[ {xf\left( x \right)} \right]} \right|_{x = 3}} = 9 \cr} $$
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