Tìm đạo hàm cấp hai của hàm số \(y = \frac{x}{{{x^2} - 1}}\)
Ta có:
\(\begin{array}{l}
y = \frac{{2x}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}} = \frac{{x + 1 + x - 1}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \frac{1}{{2\left( {x - 1} \right)}} + \frac{1}{{2\left( {x + 1} \right)}} = \frac{1}{2}\left( {\frac{1}{{x - 1}} + \frac{1}{{x + 1}}} \right)
\end{array}\)
Suy ra:
\(\begin{array}{l}
y' = \frac{1}{2}\left[ { - \frac{1}{{{{\left( {x - 1} \right)}^2}}} - \frac{1}{{{{\left( {x + 1} \right)}^2}}}} \right]\\
y'' = \frac{1}{2}\left[ {\frac{2}{{{{\left( {x - 1} \right)}^3}}} + \frac{2}{{{{\left( {x + 1} \right)}^3}}}} \right] = \frac{1}{{{{\left( {x - 1} \right)}^3}}} + \frac{1}{{{{\left( {x + 1} \right)}^3}}}
\end{array}\)
-- Mod Toán 11