## Calculus with Applications (10th Edition)

$$f'''\left( x \right) = \frac{{18}}{{{{\left( {x + 2} \right)}^4}}},\,\,\,\,\,\,\,\,{f^{\left( 4 \right)}}\left( x \right) = - \frac{{72}}{{{{\left( {x + 2} \right)}^5}}}$$
\eqalign{ & f\left( x \right) = \frac{{x - 1}}{{x + 2}} \cr & {\text{find the derivative of }}f\left( x \right) \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{x - 1}}{{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[ {x - 1} \right] - \left( {x - 1} \right)\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( 1 \right) - \left( {x - 1} \right)\left( 1 \right)}}{{{{\left( {x + 2} \right)}^2}}} \cr & {\text{simplifying}} \cr & f'\left( x \right) = \frac{{x + 2 - x + 1}}{{{{\left( {x + 2} \right)}^2}}} \cr & f'\left( x \right) = \frac{3}{{{{\left( {x + 2} \right)}^2}}} \cr & {\text{or}} \cr & f'\left( x \right) = 3{\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[ {3{{\left( {x + 2} \right)}^{ - 2}}} \right] \cr & {\text{by using the power rule with the chain rule}} \cr & f''\left( x \right) = 3\left( { - 2} \right){\left( {x + 2} \right)^{ - 3}}\frac{d}{{dx}}\left[ {x + 2} \right] \cr & f''\left( x \right) = - 6{\left( {x + 2} \right)^{ - 3}}\left( 1 \right) \cr & f''\left( x \right) = - 6{\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[ { - 6{{\left( {x + 2} \right)}^{ - 3}}} \right] \cr & {\text{by using the power rule with the chain rule}} \cr & f'''\left( x \right) = - 6\left( { - 3} \right){\left( {x + 2} \right)^{ - 4}}\frac{d}{{dx}}\left[ {x + 2} \right] \cr & f'''\left( x \right) = 18{\left( {x + 2} \right)^{ - 4}}\left( 1 \right) \cr & f'''\left( x \right) = 18{\left( {x + 2} \right)^{ - 4}} \cr & or \cr & f'''\left( x \right) = \frac{{18}}{{{{\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[ {18{{\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) = 18\left( { - 4} \right){\left( {x + 2} \right)^{ - 5}}\left( 1 \right) \cr & {f^{\left( 4 \right)}}\left( x \right) = - 72{\left( {x + 2} \right)^{ - 5}} \cr & or \cr & {f^{\left( 4 \right)}}\left( x \right) = - \frac{{72}}{{{{\left( {x + 2} \right)}^5}}} \cr}