## Calculus with Applications (10th Edition)

$$f''\left( x \right) = \frac{4}{{{{\left( {{x^2} + 4} \right)}^{3/2}}}},f''\left( 0 \right) = \frac{1}{2},f''\left( 2 \right) = \frac{1}{{4\sqrt 2 }}$$
\eqalign{ & f\left( x \right) = \sqrt {{x^2} + 4} \cr & {\text{write the radical of the function as }}{\left( {{x^2} + 4} \right)^{1/2}} \cr & {\text{ }}f\left( x \right) = {\left( {{x^2} + 4} \right)^{1/2}} \cr & {\text{find the derivative of }}f\left( x \right) \cr & {\text{ }}f'\left( x \right) = \frac{d}{{dx}}\left[ {{{\left( {{x^2} + 4} \right)}^{1/2}}} \right] \cr & {\text{use power rule with the chain rule}} \cr & {\text{ }}f'\left( x \right) = \frac{1}{2}{\left( {{x^2} + 4} \right)^{ - 1/2}}\frac{d}{{dx}}\left[ {{x^2} + 4} \right] \cr & {\text{ }}f'\left( x \right) = \frac{1}{2}{\left( {{x^2} + 4} \right)^{ - 1/2}}\left( {2x} \right) \cr & {\text{simplifying}} \cr & {\text{ }}f'\left( x \right) = x{\left( {{x^2} + 4} \right)^{ - 1/2}} \cr & \cr & {\text{find the derivative of }}f'\left( x \right) \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {x{{\left( {{x^2} + 4} \right)}^{ - 1/2}}} \right] \cr & {\text{by using the product rule}} \cr & f''\left( x \right) = x\frac{d}{{dx}}\left[ {{{\left( {{x^2} + 4} \right)}^{ - 1/2}}} \right] + {\left( {{x^2} + 4} \right)^{ - 1/2}}\frac{d}{{dx}}\left[ x \right] \cr & {\text{solving derivatives}} \cr & f''\left( x \right) = x\left[ { - \frac{1}{2}{{\left( {{x^2} + 4} \right)}^{ - 3/2}}\left( {2x} \right)} \right] + {\left( {{x^2} + 4} \right)^{ - 1/2}}\left( 1 \right) \cr & f''\left( x \right) = - {x^2}{\left( {{x^2} + 4} \right)^{ - 3/2}} + {\left( {{x^2} + 4} \right)^{ - 1/2}} \cr & {\text{factoring}} \cr & f''\left( x \right) = {\left( {{x^2} + 4} \right)^{ - 3/2}}\left( { - {x^2} + {x^2} + 4} \right) \cr & f''\left( x \right) = 4{\left( {{x^2} + 4} \right)^{ - 3/2}} \cr & or \cr & f''\left( x \right) = \frac{4}{{{{\left( {{x^2} + 4} \right)}^{3/2}}}} \cr & \cr & {\text{find }}f''\left( 0 \right){\text{ and }}f''\left( 2 \right) \cr & f''\left( 0 \right) = \frac{4}{{{{\left( {{0^2} + 4} \right)}^{3/2}}}} = \frac{4}{{4\sqrt 4 }} = \frac{1}{2} \cr & f''\left( 2 \right) = \frac{4}{{{{\left( {{2^2} + 4} \right)}^{3/2}}}} = \frac{4}{{8\sqrt 8 }} = \frac{1}{{2\sqrt 8 }} = \frac{1}{{4\sqrt 2 }} \cr}