Tính đạo hàm của y = ( x − 2 ) √ x^2 + 1 .

Lời giải:

\({\rm{y'}} = {\left[ {\left( {{\rm{x}} - 2} \right)\sqrt {{{\rm{x}}^2} + 1} } \right]^\prime }\)

\( = {\left( {{\rm{x}} - 2} \right)^\prime }\sqrt {{{\rm{x}}^2} + 1} + \left( {{\rm{x}} - 2} \right){\left( {\sqrt {{{\rm{x}}^2} + 1} } \right)^\prime }\)

\( = \sqrt {{{\rm{x}}^2} + 1} + \left( {{\rm{x}} - 2} \right)\frac{{{{\left( {{{\rm{x}}^2} + 1} \right)}^\prime }}}{{2\sqrt {{{\rm{x}}^2} + 1} }}\)

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

\( = \frac{{1 + {{\rm{x}}^2} + {{\rm{x}}^2} - 2{\rm{x}}}}{{\sqrt {{{\rm{x}}^2} + 1} }}\)

\( = \frac{{2{{\rm{x}}^2} - 2{\rm{x}} + 1}}{{\sqrt {{{\rm{x}}^2} + 1} }}\).