Bài tập 5.26 trang 203 SBT Toán 11

Ta có: 

\(\varphi \left( x \right) = \frac{{\left( {x - 2} \right)\left( {8 - x} \right)}}{{{x^2}}} = \frac{{ - {x^2} + 10x - 16}}{{{x^2}}} =  - 1 + \frac{{10}}{x} - \frac{{16}}{{{x^2}}} \Rightarrow \varphi '\left( x \right) =  - \frac{{10}}{{{x^2}}} + \frac{{32}}{{{x^3}}}\)

Vậy \(\varphi '(2) =  - \frac{{10}}{4} + \frac{{32}}{8} = \frac{3}{2}\)