Hướng dẫn giải:

Chọn D.

Áp dụng \({\left( {{u^\alpha }} \right)^/},\) với \(u = \cos \left( {\frac{{\sqrt x + 1}}{{\sqrt x - 1}}} \right)\)

\(y' = 2.\cos \left( {\frac{{\sqrt x + 1}}{{\sqrt x - 1}}} \right).{\left[ {\cos \left( {\frac{{\sqrt x + 1}}{{\sqrt x - 1}}} \right)} \right]^/} = - 2.\cos \left( {\frac{{\sqrt x + 1}}{{\sqrt x - 1}}} \right).\sin \left( {\frac{{\sqrt x + 1}}{{\sqrt x - 1}}} \right).{\left( {\frac{{\sqrt x + 1}}{{\sqrt x - 1}}} \right)^/}\)

\(y' = - \sin \left( {2\frac{{\sqrt x + 1}}{{\sqrt x - 1}}} \right).{\left( {\frac{{\sqrt x + 1}}{{\sqrt x - 1}}} \right)^/}.\)

Tính \({\left( {\frac{{\sqrt x + 1}}{{\sqrt x - 1}}} \right)^/} = \frac{{{{\left( {\sqrt x + 1} \right)}^/}.\left( {\sqrt x - 1} \right) - {{\left( {\sqrt x - 1} \right)}^/}.\left( {\sqrt x + 1} \right)}}{{{{\left( {\sqrt x - 1} \right)}^2}}} = \frac{{ - 1}}{{\sqrt x {{\left( {\sqrt x - 1} \right)}^2}}}.\)

Vậy \(y' = \frac{1}{{\sqrt x {{\left( {\sqrt x - 1} \right)}^2}}}.\sin \left( {2.\frac{{\sqrt x + 1}}{{\sqrt x - 1}}} \right).\)