Answer
$$\frac{{dy}}{{dx}} = \sqrt {\frac{{x - 1}}{{{x^2} + 2}}} \left( {\frac{1}{{2\left( {x - 1} \right)}} - \frac{x}{{{x^2} + 1}}} \right)$$
Work Step by Step
$$\eqalign{
& y = \sqrt {\frac{{x - 1}}{{{x^2} + 2}}} \cr
& {\text{Rewrite the function using the radical properties}} \cr
& y = \frac{{\sqrt {x - 1} }}{{\sqrt {{x^2} + 2} }} \cr
& y = \frac{{{{\left( {x - 1} \right)}^{1/2}}}}{{{{\left( {{x^2} + 2} \right)}^{1/2}}}} \cr
& {\text{Differentiate by logarithmic differentiation}} \cr
& {\text{Take the natural logarithm of both sides}} \cr
& \ln y = \ln \frac{{{{\left( {x - 1} \right)}^{1/2}}}}{{{{\left( {{x^2} + 2} \right)}^{1/2}}}} \cr
& {\text{Use the property }}\ln \left( {\frac{a}{b}} \right) = \ln a - \ln b \cr
& \ln y = \ln {\left( {x - 1} \right)^{1/2}} + \ln {\left( {{x^2} + 2} \right)^{1/2}} \cr
& {\text{Use the property }}\ln \left( {{a^n}} \right) = n\ln a \cr
& \ln y = \frac{1}{2}\ln \left( {x - 1} \right) - \frac{1}{2}\ln \left( {{x^2} + 2} \right) \cr
& {\text{Differentiate both sides w}}{\text{.r}}{\text{.t}} \hspace{2mm} x \cr
& \frac{d}{{dx}}\left[ {\ln y} \right] = \frac{1}{2}\frac{d}{{dx}}\left[ {\ln \left( {x - 1} \right)} \right] - \frac{1}{2}\frac{d}{{dx}}\left[ {\ln \left( {{x^2} + 2} \right)} \right] \cr
& \frac{d}{{dx}}\left[ {\ln y} \right] = \frac{1}{2}\left( {\frac{1}{{x - 1}}} \right) - \frac{1}{2}\left( {\frac{{2x}}{{{x^2} + 1}}} \right) \cr
& \frac{1}{y}\frac{{dy}}{{dx}} = \frac{1}{{2\left( {x - 1} \right)}} - \frac{x}{{{x^2} + 1}} \cr
& \frac{{dy}}{{dx}} = y\left( {\frac{1}{{2\left( {x - 1} \right)}} - \frac{x}{{{x^2} + 1}}} \right) \cr
& {\text{Substitute back }}y = \sqrt {\frac{{x - 1}}{{{x^2} + 2}}} \cr
& \frac{{dy}}{{dx}} = \sqrt {\frac{{x - 1}}{{{x^2} + 2}}} \left( {\frac{1}{{2\left( {x - 1} \right)}} - \frac{x}{{{x^2} + 1}}} \right) \cr} $$
