Answer
$$\eqalign{
& f'\left( x \right) = {\left( {{x^2} + 1} \right)^{x - 1}}\left[ {2{x^2} + \left( {{x^2} + 1} \right)\ln \left( {{x^2} + 1} \right)} \right] \cr
& f'\left( 1 \right) = 2\left( {1 + \ln 2} \right) \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {\left( {{x^2} + 1} \right)^x};\,\,\,\,\,\,a = 1 \cr
& {\text{Use }}{b^x} = {e^{x\ln b}},{\text{ then}} \cr
& f\left( x \right) = {e^{x\ln \left( {{x^2} + 1} \right)}} \cr
& {\text{Differentiating}} \cr
& f'\left( x \right) = {e^{x\ln \left( {{x^2} + 1} \right)}}\frac{d}{{dx}}\left[ {x\ln \left( {{x^2} + 1} \right)} \right] \cr
& f'\left( x \right) = {\left( {{x^2} + 1} \right)^x}\left[ {x\left( {\frac{{2x}}{{{x^2} + 1}}} \right) + \ln \left( {{x^2} + 1} \right)} \right] \cr
& f'\left( x \right) = {\left( {{x^2} + 1} \right)^x}\left[ {\frac{{2{x^2} + \left( {{x^2} + 1} \right)\ln \left( {{x^2} + 1} \right)}}{{{x^2} + 1}}} \right] \cr
& f'\left( x \right) = {\left( {{x^2} + 1} \right)^{x - 1}}\left[ {2{x^2} + \left( {{x^2} + 1} \right)\ln \left( {{x^2} + 1} \right)} \right] \cr
& \cr
& {\text{Evaluate the derivative at }}x = 1 \cr
& f'\left( 1 \right) = {\left( {{1^2} + 1} \right)^{1 - 1}}\left[ {2{{\left( 1 \right)}^2} + \left( {{1^2} + 1} \right)\ln \left( {{1^2} + 1} \right)} \right] \cr
& f'\left( 1 \right) = 1\left[ {2 + \left( 2 \right)\ln \left( 2 \right)} \right] \cr
& f'\left( 1 \right) = 2 + 2\ln 2 \cr
& f'\left( 1 \right) = 2\left( {1 + \ln 2} \right) \cr} $$
