Ta có:
$\left(\dfrac{u(x)}{v(x)}\right)' = \dfrac{u'(x).v(x) - v'(x).u(x)}{[v(x)]^2}$
Áp dụng:
$y = \dfrac{x^2 + 2x + 2}{x +1}\qquad (x \ne -1)$
$\to y' = \left(\dfrac{x^2 + 2x + 2}{x +1}\right)'$
$\to y' = \dfrac{(x^2 + 2x +2)'.(x +1) - (x +1)'(x^2 + 2x + 2)}{(x +1)^2}$
$\to y' = \dfrac{(2x + 2)(x +1) - 1.(x^2 + 2x + 2)}{(x +1)^2}$
$\to y' = \dfrac{(2x^2 + 4x + 2) - x^2 - 2x - 2}{(x +1)^2}$
$\to y' = \dfrac{x^2 + 2x}{(x +1)^2}$