Chapter 5 - Graphs and the Derivative - 5.3 Higher Derivatives, Concavity, and the Second Derivative Test - 5.3 Exercises - Page 283: 10

$$f''\left( x \right) = \frac{{18}}{{\left( {2{x^2} + 9} \right)\sqrt {2{x^2} + 9} }},f''\left( 0 \right) = \frac{2}{3},f''\left( 2 \right) = \frac{{18}}{{17\sqrt {17} }}$$

$$\eqalign{ & f\left( x \right) = \sqrt {2{x^2} + 9} \cr & {\text{write the radical of the function as }}{\left( {{x^2} + 4} \right)^{1/2}} \cr & {\text{ }}f\left( x \right) = {\left( {2{x^2} + 9} \right)^{1/2}} \cr & {\text{find the derivative of }}f\left( x \right) \cr & {\text{ }}f'\left( x \right) = \frac{d}{{dx}}\left[ {{{\left( {2{x^2} + 9} \right)}^{1/2}}} \right] \cr & {\text{use power rule with the chain rule}} \cr & {\text{ }}f'\left( x \right) = \frac{1}{2}{\left( {2{x^2} + 9} \right)^{ - 1/2}}\frac{d}{{dx}}\left[ {2{x^2} + 9} \right] \cr & {\text{ }}f'\left( x \right) = \frac{1}{2}{\left( {2{x^2} + 9} \right)^{ - 1/2}}\left( {4x} \right) \cr & {\text{simplifying}} \cr & {\text{ }}f'\left( x \right) = 2x{\left( {2{x^2} + 9} \right)^{ - 1/2}} \cr & \cr & {\text{find the derivative of }}f'\left( x \right) \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {2x{{\left( {2{x^2} + 9} \right)}^{ - 1/2}}} \right] \cr & {\text{by using the product rule}} \cr & f''\left( x \right) = 2x\frac{d}{{dx}}\left[ {{{\left( {2{x^2} + 9} \right)}^{ - 1/2}}} \right] + {\left( {2{x^2} + 9} \right)^{ - 1/2}}\frac{d}{{dx}}\left[ {2x} \right] \cr & {\text{solving derivatives}} \cr & f''\left( x \right) = 2x\left[ { - \frac{1}{2}{{\left( {2{x^2} + 9} \right)}^{ - 3/2}}\left( {4x} \right)} \right] + {\left( {2{x^2} + 9} \right)^{ - 1/2}}\left( 2 \right) \cr & f''\left( x \right) = - 4{x^2}{\left( {2{x^2} + 9} \right)^{ - 3/2}} + 2{\left( {2{x^2} + 9} \right)^{ - 1/2}} \cr & {\text{factoring}} \cr & f''\left( x \right) = 2{\left( {2{x^2} + 9} \right)^{ - 3/2}}\left( { - 2{x^2} + 2{x^2} + 9} \right) \cr & f''\left( x \right) = 2{\left( {2{x^2} + 9} \right)^{ - 3/2}}\left( 9 \right) \cr & f''\left( x \right) = 18{\left( {2{x^2} + 9} \right)^{ - 3/2}} \cr & or \cr & f''\left( x \right) = \frac{{18}}{{{{\left( {2{x^2} + 9} \right)}^{3/2}}}} \cr & f''\left( x \right) = \frac{{18}}{{\left( {2{x^2} + 9} \right)\sqrt {2{x^2} + 9} }} \cr & \cr & {\text{find }}f''\left( 0 \right){\text{ and }}f''\left( 2 \right) \cr & f''\left( 0 \right) = \frac{{18}}{{\left( {2{{\left( 0 \right)}^2} + 9} \right)\sqrt {2{{\left( 0 \right)}^2} + 9} }} = \frac{2}{3} \cr & f''\left( 0 \right) = \frac{{18}}{{\left( {2{{\left( 2 \right)}^2} + 9} \right)\sqrt {2{{\left( 2 \right)}^2} + 9} }} = \frac{{18}}{{17\sqrt {17} }} \cr} $$