Chapter 5 - Graphs and the Derivative - 5.3 Higher Derivatives, Concavity, and the Second Derivative Test - 5.3 Exercises - Page 283: 15

$$f''\left( x \right) = \frac{{2\ln x - 3}}{{4{x^3}}},\,\,\,\,\,f''\left( 0 \right) = {\text{undefined}},\,\,\,\,\,f''\left( 2 \right) = \frac{{2\ln 2 - 3}}{{32}}$$

$$\eqalign{ & f\left( x \right) = \frac{{\ln x}}{{4x}} \cr & {\text{find the derivative of }}f\left( x \right) \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{\ln x}}{{4x}}} \right] \cr & {\text{by using the quotient rule }}\frac{d}{{dx}}\left[ {\frac{{u\left( x \right)}}{{v\left( x \right)}}} \right] = \frac{{v\left( x \right) \cdot u'\left( x \right) - u\left( x \right)v'\left( x \right)}}{{{{\left[ {v\left( x \right)} \right]}^2}}} \cr & f'\left( x \right) = \frac{{\left( {4x} \right)\frac{d}{{dx}}\left[ {\ln x} \right] - \ln x\frac{d}{{dx}}\left[ {4x} \right]}}{{{{\left( {4x} \right)}^2}}} \cr & {\text{solve derivatives}} \cr & f'\left( x \right) = \frac{{\left( {4x} \right)\left( {\frac{1}{x}} \right) - \ln x\left( 4 \right)}}{{{{\left( {4x} \right)}^2}}} \cr & {\text{simplifying}} \cr & f'\left( x \right) = \frac{{4 - 4\ln x}}{{16{x^2}}} \cr & f'\left( x \right) = \frac{{1 - \ln x}}{{4{x^2}}} \cr & {\text{find the derivative of }}f'\left( x \right) \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{1 - \ln x}}{{4{x^2}}}} \right] \cr & {\text{by using the quotient rule}} \cr & f''\left( x \right) = \frac{{\left( {4{x^2}} \right)\left( { - \frac{1}{x}} \right) - \left( {1 - \ln x} \right)\left( {8x} \right)}}{{{{\left( {4{x^2}} \right)}^2}}} \cr & f''\left( x \right) = \frac{{ - 4x - 8x + 8x\ln x}}{{16{x^4}}} \cr & {\text{simplifying}} \cr & f''\left( x \right) = \frac{{8x\ln x - 12x}}{{16{x^4}}} \cr & f''\left( x \right) = \frac{{2\ln x - 3}}{{4{x^3}}} \cr & \cr & {\text{find }}f''\left( 0 \right){\text{ and }}f''\left( 2 \right) \cr & f''\left( 0 \right) = \frac{{2\ln \left( 0 \right) - 3}}{{4{{\left( 0 \right)}^3}}},{\text{ The denominator cannot be 0 and }}\ln x{\text{ is not defined for }}o \cr & {\text{then the second derivative is not defined for }}x = 0 \cr & f''\left( 2 \right) = \frac{{2\ln \left( 2 \right) - 3}}{{4{{\left( 2 \right)}^3}}} = \frac{{2\ln 2 - 3}}{{32}} \cr} $$