Cho hàm số \(f\left( x \right) = + 2}}{{x - 2}}\) . Giá trị 1 \right)\) bằng

* Hướng dẫn giải

Ta có \(f\left( x \right) = \frac{{{x^2} + 2}}{{x - 2}}\)  (ĐK: \(x \ne 2\))

Suy ra \(f'\left( x \right) = {\left( {\frac{{{x^2} + 2}}{{x - 2}}} \right)^\prime }\)

\(\begin{array}{l} = \frac{{{{\left( {{x^2} + 2} \right)}^\prime }\left( {x - 2} \right) - \left( {{x^2} + 2} \right){{\left( {x - 2} \right)}^\prime }}}{{{{\left( {x - 2} \right)}^2}}}\\ = \frac{{2x\left( {x - 2} \right) - \left( {{x^2} + 2} \right)}}{{{{\left( {x - 2} \right)}^2}}}\end{array}\)

\( = \frac{{2{x^2} - 4x - {x^2} - 2}}{{{{\left( {x - 2} \right)}^2}}} = \frac{{{x^2} - 4x - 2}}{{{{\left( {x - 2} \right)}^2}}}\)

Suy ra \(f'\left( 1 \right) = \frac{{{1^2} - 4.1 - 2}}{{{{\left( {1 - 2} \right)}^2}}} =  - 5\)