#### Answer

$$\frac{{{d^2}\left( {{{\log }_{10}}x} \right)}}{{d{x^2}}} = - \frac{1}{{{x^2}\left( {\ln 10} \right)}}$$

#### Work Step by Step

$$\eqalign{
& {\text{let }}y = {\log _{10}}x \cr
& {\text{find }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{{\log }_{10}}x} \right] \cr
& {\text{using the rule }}\frac{d}{{dx}}\left[ {{{\log }_a}x} \right] = \frac{1}{{u\ln a}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{x\ln 10}} \cr
& \cr
& {\text{find }}\frac{{{d^2}y}}{{d{x^2}}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left[ {\frac{{dy}}{{dx}}} \right] \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left[ {\frac{1}{{x\ln 10}}} \right] \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{1}{{\ln 10}}\frac{d}{{dx}}\left[ {\frac{1}{x}} \right] \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{1}{{\ln 10}}\left( { - \frac{1}{{{x^2}}}} \right) \cr
& \frac{{{d^2}y}}{{d{x^2}}} = - \frac{1}{{{x^2}\left( {\ln 10} \right)}} \cr
& {\text{where }}y = {\log _{10}}x \cr
& \frac{{{d^2}\left( {{{\log }_{10}}x} \right)}}{{d{x^2}}} = - \frac{1}{{{x^2}\left( {\ln 10} \right)}} \cr} $$