Answer
$$\left( {\bf{a}} \right)\frac{d}{{dx}}\left[ {{{\log }_x}e} \right] = - \frac{1}{{x{{\ln }^2}x}},\left( {\bf{b}} \right)\frac{d}{{dx}}\left[ {{{\log }_x}e} \right] = - \frac{{\ln 2}}{{x{{\ln }^2}x}}$$
Work Step by Step
$$\eqalign{
& \left( {\bf{a}} \right)\frac{d}{{dx}}\left[ {{{\log }_x}e} \right] \cr
& {\text{where }}{\log _a}b = \frac{{\ln b}}{{\ln a}}.{\text{ Thus}}{\text{,}} \cr
& \frac{d}{{dx}}\left[ {{{\log }_x}e} \right] = \frac{d}{{dx}}\left[ {\frac{{\ln e}}{{\ln x}}} \right] \cr
& \frac{d}{{dx}}\left[ {{{\log }_x}e} \right] = \frac{d}{{dx}}\left[ {\frac{1}{{\ln x}}} \right] \cr
& \frac{d}{{dx}}\left[ {{{\log }_x}e} \right] = \frac{d}{{dx}}\left[ {{{\left( {\ln x} \right)}^{ - 1}}} \right] \cr
& {\text{By the chain rule}} \cr
& \frac{d}{{dx}}\left[ {{{\log }_x}e} \right] = - {\left( {\ln x} \right)^{ - 2}}\frac{d}{{dx}}\left[ {\ln x} \right] \cr
& \frac{d}{{dx}}\left[ {{{\log }_x}e} \right] = - {\left( {\ln x} \right)^{ - 2}}\left( {\frac{1}{x}} \right) \cr
& {\text{Simplify}} \cr
& \frac{d}{{dx}}\left[ {{{\log }_x}e} \right] = - \frac{1}{{{{\left( {\ln x} \right)}^2}}}\left( {\frac{1}{x}} \right) \cr
& \frac{d}{{dx}}\left[ {{{\log }_x}e} \right] = - \frac{1}{{x{{\ln }^2}x}} \cr
& \cr
& \left( {\bf{b}} \right) \cr
& {\text{where }}{\log _a}b = \frac{{\ln b}}{{\ln a}}.{\text{ Thus}}{\text{,}} \cr
& \frac{d}{{dx}}\left[ {{{\log }_x}2} \right] = \frac{d}{{dx}}\left[ {\frac{{\ln 2}}{{\ln x}}} \right] \cr
& \frac{d}{{dx}}\left[ {{{\log }_x}2} \right] = \ln 2\frac{d}{{dx}}\left[ {\frac{1}{{\ln x}}} \right] \cr
& \frac{d}{{dx}}\left[ {{{\log }_x}2} \right] = \ln 2\frac{d}{{dx}}\left[ {{{\left( {\ln x} \right)}^{ - 1}}} \right] \cr
& {\text{By the chain rule}} \cr
& \frac{d}{{dx}}\left[ {{{\log }_x}2} \right] = - \ln 2{\left( {\ln x} \right)^{ - 2}}\frac{d}{{dx}}\left[ {\ln x} \right] \cr
& \frac{d}{{dx}}\left[ {{{\log }_x}2} \right] = - \ln 2{\left( {\ln x} \right)^{ - 2}}\left( {\frac{1}{x}} \right) \cr
& {\text{Simplify}} \cr
& \frac{d}{{dx}}\left[ {{{\log }_x}2} \right] = - \frac{{\ln 2}}{{{{\left( {\ln x} \right)}^2}}}\left( {\frac{1}{x}} \right) \cr
& \frac{d}{{dx}}\left[ {{{\log }_x}e} \right] = - \frac{{\ln 2}}{{x{{\ln }^2}x}} \cr} $$
