Tìm
biết \(y = \frac{{{x^2}}}{{1 - x}}\)
\(\frac{{ - 2}}{{{{\left( {1 - x} \right)}^3}}}\)
C. \(\frac{2}{{{{\left( {1 - x} \right)}^3}}}\)
Ta có:
\(\begin{array}{l}
y = \frac{{{x^2}}}{{1 - x}} = \frac{{{x^2} - 1 + 1}}{{1 - x}} = x + 1 + \frac{1}{{1 - x}}\\
\Rightarrow y' = 1 + \frac{1}{{{{\left( {1 - x} \right)}^2}}}\\
\Rightarrow y'' = \frac{2}{{{{\left( {1 - x} \right)}^3}}}
\end{array}\)
Chọn C.
-- Mod Toán 11