Answer
$$\frac{d}{{dx}}{\left( {1 + \frac{1}{x}} \right)^x} = {\left( {1 + \frac{1}{x}} \right)^x}\left( {\ln \left( {\frac{{x + 1}}{x}} \right) - \frac{1}{{x + 1}}} \right)$$
Work Step by Step
$$\eqalign{
& \frac{d}{{dx}}{\left( {1 + \frac{1}{x}} \right)^x} \cr
& or \cr
& \frac{d}{{dx}}{\left( {\frac{{x + 1}}{x}} \right)^x} \cr
& {\text{let }}y = {\left( {\frac{{x + 1}}{x}} \right)^x} \cr
& {\text{taking the natural logarithm of both sides of the equation}} \cr
& \,\,\,\ln y = \ln {\left( {\frac{{x + 1}}{x}} \right)^x} \cr
& {\text{using logarithmic properties}} \cr
& \,\,\,\ln y = x\ln \left( {\frac{{x + 1}}{x}} \right) \cr
& \,\,\,\ln y = x\ln \left( {x + 1} \right) - x\ln x \cr
& {\text{differentiate both sides with respect to }}x \cr
& \frac{d}{{dx}}\left[ {\ln y} \right] = \frac{d}{{dx}}\left[ {x\ln \left( {x + 1} \right)} \right] - \frac{d}{{dx}}\left[ {x\ln \left( x \right)} \right] \cr
& {\text{compute derivatives}} \cr
& \frac{1}{y}\frac{{dy}}{{dx}} = x\frac{d}{{dx}}\left[ {\ln \left( {x + 1} \right)} \right] + \ln \left( {x + 1} \right)\frac{d}{{dx}}\left[ x \right] - x\frac{d}{{dx}}\left[ {\ln \left( x \right)} \right] - \ln \left( x \right)\frac{d}{{dx}}\left[ x \right] \cr
& \frac{1}{y}\frac{{dy}}{{dx}} = x\left( {\frac{1}{{x + 1}}} \right) + \ln \left( {x + 1} \right) - x\left( {\frac{1}{x}} \right) - \ln \left( x \right) \cr
& \frac{1}{y}\frac{{dy}}{{dx}} = \frac{x}{{x + 1}} + \ln \left( {x + 1} \right) - 1 - \ln \left( x \right) \cr
& \frac{1}{y}\frac{{dy}}{{dx}} = \frac{x}{{x + 1}} + \ln \left( {\frac{{x + 1}}{x}} \right) - 1 + \cr
& \frac{1}{y}\frac{{dy}}{{dx}} = \ln \left( {\frac{{x + 1}}{x}} \right) - \frac{1}{{x + 1}} \cr
& {\text{solving for }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = y\left( {\ln \left( {\frac{{x + 1}}{x}} \right) - \frac{1}{{x + 1}}} \right) \cr
& {\text{replace }}y{\text{ with }}{\left( {1 + \frac{1}{x}} \right)^x} \cr
& \frac{d}{{dx}}{\left( {1 + \frac{1}{x}} \right)^x} = {\left( {1 + \frac{1}{x}} \right)^x}\left( {\ln \left( {\frac{{x + 1}}{x}} \right) - \frac{1}{{x + 1}}} \right) \cr} $$
