Chapter 5 - Graphs and the Derivative - Chapter Review - Review Exercises - Page 298: 37

$${\text{ }}f''\left( t \right) = \frac{1}{{{{\left( {{t^2} + 1} \right)}^{3/2}}}},\,\,\,\,\,f''\left( 1 \right) = \frac{1}{{2\sqrt 2 }},\,\,\,\,f''\left( { - 3} \right) = \frac{1}{{10\sqrt {10} }}$$

Work Step by Step

\eqalign{ & f\left( t \right) = \sqrt {{t^2} + 1} \cr & {\text{write the radical }}\sqrt {{t^2} + 1} {\text{ as }}{\left( {{t^2} + 1} \right)^{1/2}} \cr & f\left( t \right) = {\left( {{t^2} + 1} \right)^{1/2}} \cr & {\text{find the derivative of }}f'\left( t \right) \cr & f'\left( t \right) = \frac{d}{{dt}}\left[ {{{\left( {{t^2} + 1} \right)}^{1/2}}} \right] \cr & {\text{by using the power rule with the chain rule}} \cr & f'\left( t \right) = \frac{1}{2}{\left( {{t^2} + 1} \right)^{ - 1/2}}\frac{d}{{dt}}\left[ {{t^2} + 1} \right] \cr & f'\left( t \right) = \frac{1}{2}{\left( {{t^2} + 1} \right)^{ - 1/2}}\left( {2t} \right) \cr & f'\left( t \right) = t{\left( {{t^2} + 1} \right)^{ - 1/2}} \cr & \cr & {\text{find the derivative of }}f'\left( t \right) \cr & f''\left( t \right) = \frac{d}{{dt}}\left[ {t{{\left( {{t^2} + 1} \right)}^{ - 1/2}}} \right] \cr & {\text{using the product rule}} \cr & f''\left( t \right) = t\frac{d}{{dt}}\left[ {{{\left( {{t^2} + 1} \right)}^{ - 1/2}}} \right] + {\left( {{t^2} + 1} \right)^{ - 1/2}}\frac{d}{{dt}}\left[ t \right] \cr & {\text{by using the power rule with the chain rule}} \cr & f''\left( t \right) = t\left( { - \frac{1}{2}} \right){\left( {{t^2} + 1} \right)^{ - 3/2}}\frac{d}{{dt}}\left[ {{t^2} + 1} \right] + {\left( {{t^2} + 1} \right)^{ - 1/2}}\left( 1 \right) \cr & f''\left( t \right) = t\left( { - \frac{1}{2}} \right){\left( {{t^2} + 1} \right)^{ - 3/2}}\left( {2t} \right) + {\left( {{t^2} + 1} \right)^{ - 1/2}} \cr & f''\left( t \right) = - {t^2}{\left( {{t^2} + 1} \right)^{ - 3/2}} + {\left( {{t^2} + 1} \right)^{ - 1/2}} \cr & {\text{factoring}} \cr & f''\left( t \right) = {\left( {{t^2} + 1} \right)^{ - 3/2}}\left[ { - {t^2} + {t^2} + 1} \right] \cr & f''\left( t \right) = {\left( {{t^2} + 1} \right)^{ - 3/2}} \cr & f''\left( t \right) = \frac{1}{{{{\left( {{t^2} + 1} \right)}^{3/2}}}} \cr & \cr & {\text{find }}f''\left( 1 \right){\text{ and }}f''\left( { - 3} \right) \cr & f''\left( 1 \right) = \frac{1}{{{{\left( {{{\left( 1 \right)}^2} + 1} \right)}^{3/2}}}} = \frac{1}{{2\sqrt 2 }} \cr & and \cr & f''\left( { - 3} \right) = \frac{1}{{{{\left( {{{\left( { - 3} \right)}^2} + 1} \right)}^{3/2}}}} = \frac{1}{{10\sqrt {10} }} \cr}

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