Calculus with Applications (10th Edition)

Published by Pearson

Chapter 5 - Graphs and the Derivative - 5.3 Higher Derivatives, Concavity, and the Second Derivative Test - 5.3 Exercises - Page 283: 23

Answer

$$f'''\left( x \right) = - \frac{{36}}{{{{\left( {x - 2} \right)}^4}}},\,\,\,\,\,\,\,\,{f^{\left( 4 \right)}}\left( x \right) = \frac{{144}}{{{{\left( {x - 2} \right)}^5}}}$$

Work Step by Step

\eqalign{ & f\left( x \right) = \frac{{3x}}{{x - 2}} \cr & {\text{find the derivative of }}f\left( x \right) \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{3x}}{{x - 2}}} \right] \cr & {\text{by using the quotient rule }}\frac{d}{{dx}}\left[ {\frac{{u\left( x \right)}}{{v\left( x \right)}}} \right] = \frac{{v\left( x \right) \cdot u'\left( x \right) - u\left( x \right)v'\left( x \right)}}{{{{\left[ {v\left( x \right)} \right]}^2}}} \cr & f'\left( x \right) = \frac{{\left( {x - 2} \right)\frac{d}{{dx}}\left[ {3x} \right] - 3x\frac{d}{{dx}}\left[ {x - 2} \right]}}{{{{\left( {x - 2} \right)}^2}}} \cr & {\text{solve derivatives}} \cr & f'\left( x \right) = \frac{{\left( {x - 2} \right)\left( 3 \right) - 3x\left( 1 \right)}}{{{{\left( {x - 2} \right)}^2}}} \cr & {\text{simplifying}} \cr & f'\left( x \right) = \frac{{3x - 6 - 3x}}{{{{\left( {x - 2} \right)}^2}}} \cr & f'\left( x \right) = - \frac{6}{{{{\left( {x - 2} \right)}^2}}} \cr & {\text{or}} \cr & f'\left( x \right) = - 6{\left( {x - 2} \right)^{ - 2}} \cr & \cr & {\text{find the derivative of }}f'\left( x \right) \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ { - 6{{\left( {x - 2} \right)}^{ - 2}}} \right] \cr & {\text{by using the power rule with the chain rule}} \cr & f''\left( x \right) = - 6\left( { - 2} \right){\left( {x - 2} \right)^{ - 3}}\frac{d}{{dx}}\left[ {x - 2} \right] \cr & f''\left( x \right) = 12{\left( {x - 2} \right)^{ - 3}}\left( 1 \right) \cr & f''\left( x \right) = 12{\left( {x - 2} \right)^{ - 3}} \cr & \cr & {\text{find the derivative of }}f''\left( x \right) \cr & f'''\left( x \right) = \frac{d}{{dx}}\left[ {12{{\left( {x - 2} \right)}^{ - 3}}} \right] \cr & {\text{by using the power rule with the chain rule}} \cr & f'''\left( x \right) = 12\left( { - 3} \right){\left( {x - 2} \right)^{ - 4}}\frac{d}{{dx}}\left[ {x - 2} \right] \cr & f'''\left( x \right) = 12\left( { - 3} \right){\left( {x - 2} \right)^{ - 4}}\left( 1 \right) \cr & f'''\left( x \right) = - 36{\left( {x - 2} \right)^{ - 4}} \cr & or \cr & f'''\left( x \right) = - \frac{{36}}{{{{\left( {x - 2} \right)}^4}}} \cr & \cr & {\text{find the derivative of }}f'''\left( x \right) \cr & {f^{\left( 4 \right)}}\left( x \right) = \frac{d}{{dx}}\left[ { - 36{{\left( {x - 2} \right)}^{ - 4}}} \right] \cr & {\text{by using the power rule with the chain rule}} \cr & {f^{\left( 4 \right)}}\left( x \right) = - 36\left( { - 4} \right){\left( {x - 2} \right)^{ - 5}}\left( 1 \right) \cr & {f^{\left( 4 \right)}}\left( x \right) = 144{\left( {x - 2} \right)^{ - 5}} \cr & or \cr & {f^{\left( 4 \right)}}\left( x \right) = \frac{{144}}{{{{\left( {x - 2} \right)}^5}}} \cr}

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