Answer
$$\frac{{dy}}{{dx}} = 2{x^2}{\left( {{x^2} + 1} \right)^{x - 1}} + {\left( {{x^2} + 1} \right)^x}\ln \left( {{x^2} + 1} \right)$$
Work Step by Step
$$\eqalign{
& y = {\left( {{x^2} + 1} \right)^x} \cr
& \left( i \right){\text{Using the fact that }}{b^x} = {e^{x\ln b}},{\text{ then}} \cr
& y = {e^{x\ln \left( {{x^2} + 1} \right)}} \cr
& {\text{Differentiating}} \cr
& \frac{{dy}}{{dx}} = {e^{x\ln \left( {{x^2} + 1} \right)}}\frac{d}{{dx}}\left[ {x\ln \left( {{x^2} + 1} \right)} \right] \cr
& \frac{{dy}}{{dx}} = {e^{x\ln \left( {{x^2} + 1} \right)}}\left[ {x\left( {\frac{{2x}}{{{x^2} + 1}}} \right) + \ln \left( {{x^2} + 1} \right)} \right] \cr
& \frac{{dy}}{{dx}} = {e^{x\ln \left( {{x^2} + 1} \right)}}\left[ {\frac{{2{x^2}}}{{{x^2} + 1}} + \ln \left( {{x^2} + 1} \right)} \right] \cr
& {\text{Substitute }}{\left( {{x^2} + 1} \right)^x}{\text{ for }}{e^{x\ln \left( {{x^2} + 1} \right)}} \cr
& \frac{{dy}}{{dx}} = {\left( {{x^2} + 1} \right)^x}\left[ {\frac{{2{x^2}}}{{{x^2} + 1}} + \ln \left( {{x^2} + 1} \right)} \right] \cr
& {\text{Simplifying}} \cr
& \frac{{dy}}{{dx}} = 2{x^2}{\left( {{x^2} + 1} \right)^{x - 1}} + {\left( {{x^2} + 1} \right)^x}\ln \left( {{x^2} + 1} \right) \cr
& \cr
& \left( {ii} \right){\text{Using the logarithmic differentiation}} \cr
& \ln y = \ln {\left( {{x^2} + 1} \right)^x} \cr
& \ln y = x\ln \left( {{x^2} + 1} \right) \cr
& {\text{Differentiate both sides}} \cr
& \frac{1}{y}\frac{{dy}}{{dx}} = x\left( {\frac{{2x}}{{{x^2} + 1}}} \right) + \ln \left( {{x^2} + 1} \right) \cr
& \frac{1}{y}\frac{{dy}}{{dx}} = \frac{{2{x^2}}}{{{x^2} + 1}} + \ln \left( {{x^2} + 1} \right) \cr
& {\text{Solving for }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = y\frac{{2{x^2}}}{{{x^2} + 1}} + y\ln \left( {{x^2} + 1} \right) \cr
& {\text{Where }}y = {\left( {{x^2} + 1} \right)^x} \cr
& \frac{{dy}}{{dx}} = {\left( {{x^2} + 1} \right)^x}\frac{{2{x^2}}}{{{x^2} + 1}} + {\left( {{x^2} + 1} \right)^x}\ln \left( {{x^2} + 1} \right) \cr
& \frac{{dy}}{{dx}} = 2{x^2}{\left( {{x^2} + 1} \right)^{x - 1}} + {\left( {{x^2} + 1} \right)^x}\ln \left( {{x^2} + 1} \right) \cr} $$
