Đáp án:
a. \(y' = \frac{{ - {x^2} - 6x - 5}}{{{{\left( {x - 1} \right)}^4}}}\)
b. \(y' = \frac{{ - 36{x^2} - 36x - 9}}{{{{\left( {x - 1} \right)}^4}}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.y = \frac{{{{\left( {x + 1} \right)}^2}}}{{{{\left( {x - 1} \right)}^3}}}\\
y' = \frac{{2\left( {x + 1} \right){{\left( {x - 1} \right)}^3} - 3{{\left( {x - 1} \right)}^2}{{\left( {x + 1} \right)}^2}}}{{{{\left( {x - 1} \right)}^6}}}\\
= \frac{{2\left( {{x^2} - 1} \right){{\left( {x - 1} \right)}^2} - 3{{\left( {x - 1} \right)}^2}{{\left( {x + 1} \right)}^2}}}{{{{\left( {x - 1} \right)}^6}}}\\
= \frac{{{{\left( {x - 1} \right)}^2}\left[ {2\left( {{x^2} - 1} \right) - 3\left( {{x^2} + 2x + 1} \right)} \right]}}{{{{\left( {x - 1} \right)}^6}}}\\
= \frac{{2{x^2} - 2 - 3{x^2} - 6x - 3}}{{{{\left( {x - 1} \right)}^4}}}\\
= \frac{{ - {x^2} - 6x - 5}}{{{{\left( {x - 1} \right)}^4}}}\\
b.y = {\left( {\frac{{2x + 1}}{{x - 1}}} \right)^3}\\
y' = \frac{{2x - 2 - 2x - 1}}{{{{\left( {x - 1} \right)}^2}}}.3.{\left( {\frac{{2x + 1}}{{x - 1}}} \right)^2}\\
= \frac{{ - 9\left( {4{x^2} + 4x + 1} \right)}}{{{{\left( {x - 1} \right)}^4}}}\\
= \frac{{ - 36{x^2} - 36x - 9}}{{{{\left( {x - 1} \right)}^4}}}
\end{array}\)
