## Thomas' Calculus 13th Edition

$$\frac{{dy}}{{dx}} = \frac{{1 + \ln x + {{\ln }^2}x}}{{{{\left( {1 + \ln x} \right)}^2}}}$$
\eqalign{ & y = \frac{{x\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{{x\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[ {x\ln x} \right] - x\ln x\frac{d}{{dx}}\left[ {1 + \ln x} \right]}}{{{{\left( {1 + \ln x} \right)}^2}}} \cr & {\text{use the product rule for: }}\frac{d}{{dx}}\left[ {x\ln x} \right] \cr & \frac{{dy}}{{dx}} = \frac{{\left( {1 + \ln x} \right)\left( {x\frac{d}{{dx}}\left[ {\ln x} \right] + \ln x\frac{d}{{dx}}\left[ x \right]} \right) - x\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( {x\left( {\frac{1}{x}} \right) + \ln x\left( 1 \right)} \right) - x\ln x\left( {\frac{1}{x}} \right)}}{{{{\left( {1 + \ln x} \right)}^2}}} \cr & {\text{simplifying, we get:}} \cr & \frac{{dy}}{{dx}} = \frac{{\left( {1 + \ln x} \right)\left( {1 + \ln x} \right) - \ln x}}{{{{\left( {1 + \ln x} \right)}^2}}} \cr & \frac{{dy}}{{dx}} = \frac{{{{\left( {1 + \ln x} \right)}^2} - \ln x}}{{{{\left( {1 + \ln x} \right)}^2}}} \cr & \frac{{dy}}{{dx}} = \frac{{1 + 2\ln x + {{\ln }^2}x - \ln x}}{{{{\left( {1 + \ln x} \right)}^2}}} \cr & \frac{{dy}}{{dx}} = \frac{{1 + \ln x + {{\ln }^2}x}}{{{{\left( {1 + \ln x} \right)}^2}}} \cr}