Answer
$$f'\left( x \right) = 2{\left( {1 + \frac{1}{x}} \right)^{2x}}\left( {\ln \left( {\frac{{x + 1}}{x}} \right) - \frac{1}{{x + 1}}} \right)$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {\left( {1 + \frac{1}{x}} \right)^{2x}} \cr
& f\left( x \right) = {\left( {\frac{{x + 1}}{x}} \right)^{2x}} \cr
& {\text{taking the natural logarithm of both sides of the equation}} \cr
& \ln \left( {f\left( x \right)} \right) = \ln {\left( {\frac{{x + 1}}{x}} \right)^{2x}} \cr
& {\text{power property for logarithms}} \cr
& \ln \left( {f\left( x \right)} \right) = 2x\ln \left( {\frac{{x + 1}}{x}} \right) \cr
& {\text{quotient rule for logarithms}} \cr
& \ln \left( {f\left( x \right)} \right) = 2x\ln \left( {x + 1} \right) - 2x\ln \left( x \right) \cr
& {\text{differentiate both sides with respect to }}x \cr
& \frac{d}{{dx}}\left[ {\ln \left( {f\left( x \right)} \right)} \right] = \frac{d}{{dx}}\left[ {2x\ln \left( {x + 1} \right)} \right] - \frac{d}{{dx}}\left[ {2x\ln \left( x \right)} \right] \cr
& {\text{product rule}} \cr
& \frac{{f'\left( x \right)}}{{f\left( x \right)}} = 2x\frac{d}{{dx}}\left[ {\ln \left( {x + 1} \right)} \right] + \ln \left( {x + 1} \right)\frac{d}{{dx}}\left[ {2x} \right] - 2x\frac{d}{{dx}}\left[ {\ln \left( x \right)} \right] - \ln \left( x \right)\frac{d}{{dx}}\left[ {2x} \right] \cr
& \frac{{f'\left( x \right)}}{{f\left( x \right)}} = 2x\left( {\frac{1}{{x + 1}}} \right) + \ln \left( {x + 1} \right)\left( 2 \right) - 2x\left( {\frac{1}{x}} \right) - \ln \left( x \right)\left( 2 \right) \cr
& \frac{{f'\left( x \right)}}{{f\left( x \right)}} = \frac{{2x}}{{x + 1}} + 2\ln \left( {x + 1} \right) - 2 - 2\ln \left( x \right) \cr
& \frac{{f'\left( x \right)}}{{f\left( x \right)}} = \frac{{2x}}{{x + 1}} + 2\ln \left( {\frac{{x + 1}}{x}} \right) - 2 \cr
& \frac{{f'\left( x \right)}}{{f\left( x \right)}} = \frac{{2x - 2x - 2}}{{x + 1}} + 2\ln \left( {\frac{{x + 1}}{x}} \right) \cr
& \frac{{f'\left( x \right)}}{{f\left( x \right)}} = 2\ln \left( {\frac{{x + 1}}{x}} \right) - \frac{2}{{x + 1}} \cr
& {\text{solving for }}f'\left( x \right) \cr
& f'\left( x \right) = f\left( x \right)\left( {2\ln \left( {\frac{{x + 1}}{x}} \right) - \frac{2}{{x + 1}}} \right) \cr
& f'\left( x \right) = 2f\left( x \right)\left( {\ln \left( {\frac{{x + 1}}{x}} \right) - \frac{1}{{x + 1}}} \right) \cr
& {\text{replace }}f\left( x \right){\text{ with the original function:}} \cr
& f'\left( x \right) = 2{\left( {1 + \frac{1}{x}} \right)^{2x}}\left( {\ln \left( {\frac{{x + 1}}{x}} \right) - \frac{1}{{x + 1}}} \right) \cr} $$
