Chapter 6 - Exponential, Logarithmic, And Inverse Trigonometric Functions - 6.2 Derivatives And Integrals Involving Logarithmic Functions - Exercises Set 6.2 - Page 425: 18

$$y' = \frac{1}{{x\ln 10{{\left( {1 + \log x} \right)}^2}}}$$

$$\eqalign{ & y = \frac{{\log x}}{{1 + \log x}} \cr & {\text{quotient rule }} \cr & \left( {\frac{u}{v}} \right)' = \frac{{vu' - uv'}}{{{v^2}}} \cr & y' = \frac{{\left( {1 + \log x} \right)\left( {\log x} \right)' - \log x\left( {1 + \log x} \right)'}}{{{{\left( {1 + \log x} \right)}^2}}} \cr & {\text{so}} \cr & y' = \frac{{\left( {1 + \log x} \right)\left( {\frac{1}{{x\ln 10}}} \right) - \log x\left( {\frac{1}{{x\ln 10}}} \right)}}{{{{\left( {1 + \log x} \right)}^2}}} \cr & {\text{simplify}} \cr & y' = \frac{{\frac{1}{{x\ln 10}} + \frac{{\log x}}{{x\ln 10}} - \frac{{\log x}}{{x\ln 10}}}}{{{{\left( {1 + \log x} \right)}^2}}} \cr & y' = \frac{{\frac{1}{{x\ln 10}}}}{{{{\left( {1 + \log x} \right)}^2}}} \cr & y' = \frac{1}{{x\ln 10{{\left( {1 + \log x} \right)}^2}}} \cr} $$