Answer
$$\frac{{dy}}{{dx}} = |x-1|\sqrt {{x^2} + 1} \left( {\frac{{2x^2-x+1}}{{\left( {{x^2} + 1} \right)(x-1)}}} \right) $$
Work Step by Step
$$\eqalign{
& y = \sqrt {\left( {{x^2} + 1} \right){{\left( {x - 1} \right)}^2}} \cr
& y = {\left( {{x^2} + 1} \right)^{1/2}}\left( {x - 1} \right) \cr
& {\text{Take the natural log of both sides:}} \cr
& {\text{Use the properties of logarithms}} \cr
& \ln y = \ln {\left( {{x^2} + 1} \right)^{1/2}}\left( {x - 1} \right) \cr
& {\text{product rule:}} \cr
& \ln y = \ln {\left( {{x^2} + 1} \right)^{1/2}} + \ln \left( {x - 1} \right) \cr
& {\text{power rule:}} \cr
& \ln y = \frac{1}{2}\ln \left( {{x^2} + 1} \right) + \ln \left( {x - 1} \right) \cr
& {\text{Take derivatives of both sides with respect to }}x \cr
& \frac{1}{y}\frac{{dy}}{{dx}} = \frac{1}{2}\left( {\frac{{2x}}{{{x^2} + 1}}} \right) + \frac{1}{{x - 1}} \cr
& \frac{1}{y}\frac{{dy}}{{dx}} = \frac{x}{{{x^2} + 1}} + \frac{1}{{x - 1}} \cr
& {\text{solve for }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = y\left( {\frac{x}{{{x^2} + 1}} + \frac{1}{{x - 1}}} \right) \cr
& {\text{substitute }}\sqrt {\left( {{x^2} + 1} \right){{\left( {x - 1} \right)}^2}} {\text{ for }}y{\text{ }} \cr
& \frac{{dy}}{{dx}} = \sqrt {\left( {{x^2} + 1} \right){{\left( {x - 1} \right)}^2}} \left( {\frac{x}{{{x^2} + 1}} + \frac{1}{{x - 1}}} \right) \cr
& \frac{{dy}}{{dx}} = |x - 1|\sqrt {{x^2} + 1} \left( {\frac{{{x^2} - x + {x^2} + 1}}{{\left( {{x^2} + 1} \right)\left( {x - 1} \right)}}} \right) \cr
& \frac{{dy}}{{dx}} = |x-1|\sqrt {{x^2} + 1} \left( {\frac{{2x^2-x+1}}{{\left( {{x^2} + 1} \right)(x-1)}}} \right) \cr }$$
