Chapter 5 - Graphs and the Derivative - Chapter Review - Review Exercises - Page 298: 38

$${\text{ }}f''\left( t \right) = \frac{5}{{{{\left( {5 - {t^2}} \right)}^{3/2}}}},\,\,\,\,\,f''\left( 1 \right) = \frac{5}{8},\,\,\,\,f''\left( { - 3} \right) = {\text{Undefined}}$$

$$\eqalign{ & f\left( t \right) = - \sqrt {5 - {t^2}} \cr & {\text{write the radical }} - \sqrt {5 - {t^2}} {\text{ as /}}{\left( {5 - {t^2}} \right)^{1/2}} \cr & f\left( t \right) = - {\left( {5 - {t^2}} \right)^{1/2}} \cr & {\text{find the derivative of }}f'\left( t \right) \cr & f'\left( t \right) = - \frac{d}{{dt}}\left[ {{{\left( {5 - {t^2}} \right)}^{1/2}}} \right] \cr & {\text{by using the power rule with the chain rule}} \cr & f'\left( t \right) = - \frac{1}{2}{\left( {5 - {t^2}} \right)^{ - 1/2}}\frac{d}{{dt}}\left[ {5 - {t^2}} \right] \cr & f'\left( t \right) = - \frac{1}{2}{\left( {5 - {t^2}} \right)^{ - 1/2}}\left( { - 2t} \right) \cr & f'\left( t \right) = t{\left( {5 - {t^2}} \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( {5 - {t^2}} \right)}^{ - 1/2}}} \right] \cr & {\text{using the product rule}} \cr & f''\left( t \right) = t\frac{d}{{dt}}\left[ {{{\left( {5 - {t^2}} \right)}^{ - 1/2}}} \right] + {\left( {5 - {t^2}} \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( {5 - {t^2}} \right)^{ - 3/2}}\frac{d}{{dt}}\left[ {5 - {t^2}} \right] + {\left( {5 - {t^2}} \right)^{ - 1/2}}\left( 1 \right) \cr & f''\left( t \right) = t\left( { - \frac{1}{2}} \right){\left( {5 - {t^2}} \right)^{ - 3/2}}\left( { - 2t} \right) + {\left( {5 - {t^2}} \right)^{ - 1/2}}\left( 1 \right) \cr & f''\left( t \right) = {t^2}{\left( {5 - {t^2}} \right)^{ - 3/2}} + {\left( {5 - {t^2}} \right)^{ - 1/2}} \cr & {\text{factoring}} \cr & f''\left( t \right) = {\left( {5 - {t^2}} \right)^{ - 3/2}}\left[ {{t^2} + 5 - {t^2}} \right] \cr & f''\left( t \right) = 5{\left( {5 - {t^2}} \right)^{ - 3/2}} \cr & f''\left( t \right) = \frac{5}{{{{\left( {5 - {t^2}} \right)}^{3/2}}}} \cr & \cr & {\text{find }}f''\left( 1 \right){\text{ and }}f''\left( { - 3} \right) \cr & f''\left( 1 \right) = \frac{5}{{{{\left( {5 - {{\left( 1 \right)}^2}} \right)}^{3/2}}}} = \frac{5}{8} \cr & and \cr & f''\left( { - 3} \right) = \frac{5}{{{{\left( {5 - {{\left( { - 3} \right)}^2}} \right)}^{3/2}}}} = \frac{5}{{{{\left( { - 4} \right)}^{3/2}}}} = \frac{5}{{{{\left( {\sqrt { - 4} } \right)}^3}}} \cr & \sqrt { - 4} {\text{ is not a real number}}{\text{, so }}f''\left( t \right){\text{ is not defined for }}t = - 3 \cr} $$