Chapter 5 - Graphs and the Derivative - 5.3 Higher Derivatives, Concavity, and the Second Derivative Test - 5.3 Exercises - Page 283: 24

$$f'''\left( x \right) = \frac{{24}}{{{{\left( {2x + 1} \right)}^4}}},\,\,\,\,\,\,\,\,{f^{\left( 4 \right)}}\left( x \right) = - \frac{{192}}{{{{\left( {2x + 1} \right)}^5}}}$$

$$\eqalign{ & f\left( x \right) = \frac{x}{{2x + 1}} \cr & {\text{find the derivative of }}f\left( x \right) \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{x}{{2x + 1}}} \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( {2x + 1} \right)\frac{d}{{dx}}\left[ x \right] - x\frac{d}{{dx}}\left[ {2x + 1} \right]}}{{{{\left( {2x + 1} \right)}^2}}} \cr & {\text{solve derivatives}} \cr & f'\left( x \right) = \frac{{\left( {2x + 1} \right)\left( 1 \right) - x\frac{d}{{dx}}\left[ 2 \right]}}{{{{\left( {2x + 1} \right)}^2}}} \cr & {\text{simplifying}} \cr & f'\left( x \right) = \frac{{2x + 1 - 2x}}{{{{\left( {2x + 1} \right)}^2}}} \cr & f'\left( x \right) = \frac{1}{{{{\left( {2x + 1} \right)}^2}}} \cr & {\text{or}} \cr & f'\left( x \right) = {\left( {2x + 1} \right)^{ - 2}} \cr & \cr & {\text{find the derivative of }}f'\left( x \right) \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {{{\left( {2x + 1} \right)}^{ - 2}}} \right] \cr & {\text{by using the power rule with the chain rule}} \cr & f''\left( x \right) = - {\left( {2x + 1} \right)^{ - 3}}\frac{d}{{dx}}\left[ {2x + 1} \right] \cr & f''\left( x \right) = - {\left( {2x + 1} \right)^{ - 3}}\left( 2 \right) \cr & f''\left( x \right) = - 4{\left( {2x + 1} \right)^{ - 3}} \cr & \cr & {\text{find the derivative of }}f''\left( x \right) \cr & f'''\left( x \right) = \frac{d}{{dx}}\left[ { - 4{{\left( {2x + 1} \right)}^{ - 3}}} \right] \cr & {\text{by using the power rule with the chain rule}} \cr & f'''\left( x \right) = - 4\left( { - 3} \right){\left( {2x + 1} \right)^{ - 4}}\frac{d}{{dx}}\left[ {2x + 1} \right] \cr & f'''\left( x \right) = 12{\left( {2x + 1} \right)^{ - 4}}\left( 2 \right) \cr & f'''\left( x \right) = 24{\left( {2x + 1} \right)^{ - 4}} \cr & or \cr & f'''\left( x \right) = \frac{{24}}{{{{\left( {2x + 1} \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[ {24{{\left( {2x + 1} \right)}^{ - 4}}} \right] \cr & {\text{by using the power rule with the chain rule}} \cr & {f^{\left( 4 \right)}}\left( x \right) = 24\left( { - 4} \right){\left( {2x + 1} \right)^{ - 5}}\left( 2 \right) \cr & {f^{\left( 4 \right)}}\left( x \right) = - 192{\left( {2x + 1} \right)^{ - 5}} \cr & or \cr & {f^{\left( 4 \right)}}\left( x \right) = - \frac{{192}}{{{{\left( {2x + 1} \right)}^5}}} \cr} $$