## Thomas' Calculus 13th Edition

$$\frac{{dy}}{{dx}} = \frac{1}{{x{{\left( {1 + \ln x} \right)}^2}}}$$
\eqalign{ & y = \frac{{\ln x}}{{1 + \ln x}} \cr & {\text{Find the derivative of }}y{\text{ with respect to }}x \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\frac{{\ln x}}{{1 + \ln x}}} \right] \cr & {\text{use the quotient rule}} \cr & \frac{{dy}}{{dx}} = \frac{{\left( {1 + \ln x} \right)\frac{d}{{dx}}\left[ {\ln x} \right] - \ln x\frac{d}{{dx}}\left[ {1 + \ln x} \right]}}{{{{\left( {1 + \ln x} \right)}^2}}} \cr & {\text{solve the derivatives}} \cr & \frac{{dy}}{{dx}} = \frac{{\left( {1 + \ln x} \right)\left( {\frac{1}{x}} \right) - \ln x\left( {\frac{1}{x}} \right)}}{{{{\left( {1 + \ln x} \right)}^2}}} \cr & {\text{simplify}} \cr & \frac{{dy}}{{dx}} = \frac{{\frac{{1 + \ln x}}{x} - \frac{{\ln x}}{x}}}{{{{\left( {1 + \ln x} \right)}^2}}} \cr & \frac{{dy}}{{dx}} = \frac{{\frac{{1 + \ln x - \ln x}}{x}}}{{{{\left( {1 + \ln x} \right)}^2}}} \cr & \frac{{dy}}{{dx}} = \frac{1}{{x{{\left( {1 + \ln x} \right)}^2}}} \cr}