Áp dụng công thức tính đạo hàm ` (u/v)' = (u'v - uv')/(v^2)`

Đặt ` u = (x+1)^3 ; v = (x-1)^2`

` => u' = (x+1)^3' = 3(x+1)^2`

` v' = (x-1)^2' = 2(x-1)`

` => y' = (((x+1)^3)/((x-1)^2))' = (3(x+1)^2*(x-1)^2 - (x+1)^3*2*(x-1))/((x-1)^4)`

` = (3(x-1)^2(x+1)^2 - (x^3 +3x^2 +3x +1)*2*(x-1))/((x-1)^4)`

` = (3(x-1)^2(x+1)^2 - (2x^3 + 6x^2 +6x +2)(x-1))/(x-1)^4`

` = (3(x-1)(x+1)^2 - 2x^3 -6x^2 -6x -2)/((x-1)^3)`

` = (3*(x^2-1)(x+1) - 2x^3 -6x^2 -6x -2)/((x-1)^3)`

` = (3x^3 + 3x^2 - 3x-3 -2x^3 -6x^2 -6x -2)/((x-1)^3)`

` = (x^3 - 3x^2 - 9x - 5)/((x-1)^3)`

Đáp án:

$\begin{array}{l}
y = \dfrac{{{{\left( {x + 1} \right)}^3}}}{{{{\left( {x - 1} \right)}^2}}}\\
 \Rightarrow y' = \dfrac{{3.{{\left( {x + 1} \right)}^2}.{{\left( {x - 1} \right)}^2} - {{\left( {x + 1} \right)}^3}.2.\left( {x - 1} \right)}}{{{{\left( {x - 1} \right)}^4}}}\\
 = \dfrac{{3{{\left( {{x^2} - 1} \right)}^2} - 2.\left( {{x^3} + 3{x^2} + 3x + 1} \right)\left( {x - 1} \right)}}{{{{\left( {x - 1} \right)}^4}}}\\
 = \dfrac{{3{x^4} - 6{x^2} + 3 - 2{x^4} - 4{x^2} + 4x + 2}}{{{{\left( {x - 1} \right)}^4}}}\\
 = \dfrac{{{x^4} - 10{x^2} + 4x + 5}}{{{{\left( {x - 1} \right)}^4}}}
\end{array}$