Calculus Concepts: An Informal Approach to the Mathematics of Change 5th Edition

Published by Brooks Cole
ISBN 10: 1-43904-957-2
ISBN 13: 978-1-43904-957-0

Chapter 3 - Determining Change: Derivatives - 3.6 Activities - Page 237: 11

Answer

$$f'\left( x \right) = 15642\left( {\frac{{1 + 7.68{e^{ - 0.85x}} + 6.528x{e^{ - 0.85x}}}}{{{{\left( {1 + 7.68{e^{ - 0.85x}}} \right)}^2}}}} \right) + 1185$$

Work Step by Step

$$\eqalign{ & f\left( x \right) = \left( {79x} \right)\left( {\frac{{198}}{{1 + 7.68{e^{ - 0.85x}}}} + 15} \right) \cr & {\text{Multiply}} \cr & f\left( x \right) = \frac{{15642x}}{{1 + 7.68{e^{ - 0.85x}}}} + 1185x \cr & {\text{Calculate the derivative of the function}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left( {\frac{{15642x}}{{1 + 7.68{e^{ - 0.85x}}}}} \right) + \frac{d}{{dx}}\left( {1185x} \right) \cr & {\text{Use the quotient rule }}\left( {{\text{see the page 237}}} \right) \cr & f'\left( x \right) = \frac{{\left( {1 + 7.68{e^{ - 0.85x}}} \right)\left( {15642x} \right)' - 15642x\left( {1 + 7.68{e^{ - 0.85x}}} \right)'}}{{{{\left( {1 + 7.68{e^{ - 0.85x}}} \right)}^2}}} + 1185 \cr & {\text{Compute derivatives}} \cr & f'\left( x \right) = \frac{{\left( {1 + 7.68{e^{ - 0.85x}}} \right)\left( {15642} \right) - 15642x\left( { - 6.528{e^{ - 0.85x}}} \right)}}{{{{\left( {1 + 7.68{e^{ - 0.85x}}} \right)}^2}}} + 1185 \cr & {\text{Factor}} \cr & f'\left( x \right) = 15642\left( {\frac{{1 + 7.68{e^{ - 0.85x}} + 6.528x{e^{ - 0.85x}}}}{{{{\left( {1 + 7.68{e^{ - 0.85x}}} \right)}^2}}}} \right) + 1185 \cr} $$
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