Cho \(y = \frac{{x - 2}}{{x + 3}}\). Tìm
Ta có:
\(\begin{array}{l}
y' = \frac{{\left( {x + 3} \right) - \left( {x - 2} \right)}}{{{{\left( {x + 3} \right)}^2}}} = \frac{5}{{{{\left( {x + 3} \right)}^2}}}\\
y'' = - \frac{{10}}{{{{\left( {x + 3} \right)}^3}}}
\end{array}\)
Chọn B.
-- Mod Toán 11