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 208: 45

Answer

$$f'\left( 2 \right) = 7$$

Work Step by Step

$$\eqalign{ & {\text{let }}f\left( x \right) = \frac{1}{2}g\left( x \right) + \frac{1}{4}h\left( x \right) \cr & {\text{find the derivative}} \cr & f'\left( x \right) = {D_x}\left( {\frac{1}{2}g\left( x \right) + \frac{1}{4}h\left( x \right)} \right) \cr & {\text{use the sum rule for derivatives}} \cr & f'\left( x \right) = {D_x}\left( {\frac{1}{2}g\left( x \right)} \right) + {D_x}\left( {\frac{1}{4}h\left( x \right)} \right) \cr & {\text{use multiple constant rule for derivatives}} \cr & f'\left( x \right) = \frac{1}{2}{D_x}\left( {g\left( x \right)} \right) + \frac{1}{4}{D_x}\left( {h\left( x \right)} \right) \cr & {\text{solve derivatives}} \cr & f'\left( x \right) = \frac{1}{2}g'\left( x \right) + \frac{1}{4}h'\left( x \right) \cr & {\text{evaluate }}f'\left( 2 \right) \cr & f'\left( 2 \right) = \frac{1}{2}g'\left( 2 \right) + \frac{1}{4}h'\left( 2 \right) \cr & {\text{substitute }}g'\left( 2 \right) = 7{\text{ and }}h'\left( 2 \right) = 14 \cr & f'\left( 2 \right) = \frac{1}{2}\left( 7 \right) + \frac{1}{4}\left( {14} \right) \cr & {\text{simplifying}} \cr & f'\left( 2 \right) = \frac{7}{2} + \frac{7}{2} \cr & f'\left( 2 \right) = 7 \cr} $$
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