Tìm đạo hàm của hàm số \(y = \frac{x}{{{{\left( {1 - x} \right)}^2}{{\left( {1 + x} \right)}^3}}}\)
\(\begin{array}{l}
y = \frac{x}{{{{(1 - x)}^2}{{(1 + x)}^3}}}\\
\Rightarrow y\prime = \frac{{{{(1 - x)}^2}{{(1 + x)}^3} - x[ - 2(1 - x){{(1 + x)}^3} + {{(1 - x)}^2}.3{{(1 + x)}^2}]}}{{{{(1 - x)}^4}{{(1 + x)}^6}}}\\
= \frac{{{{(1 - x)}^2}{{(1 + x)}^3} - x(1 - {x^2})[ - 2(1 + 2x + {x^2}) + 3(1 - {x^2})]}}{{{{(1 - x)}^4}{{(1 + x)}^6}}}\\
= \frac{{{{(1 - x)}^2}{{(1 + x)}^3} - x(1 - {x^2})( - 5{x^2} - 4x + 1)}}{{{{(1 - x)}^4}{{(1 + x)}^6}}}\\
= \frac{{{{(1 - x)}^2}{{(1 + x)}^3} - x(1 - {x^2})(1 + x)(1 - 5x)}}{{{{(1 - x)}^4}{{(1 + x)}^6}}}\\
= \frac{{(1 - x){{(1 + x)}^2}[1 - {x^2} - x(1 - 5x)]}}{{{{(1 - x)}^4}{{(1 + x)}^6}}}\\
= \frac{{4{x^2} - x + 1}}{{{{(1 - x)}^4}{{(1 + x)}^6}}}
\end{array}\)
-- Mod Toán 11