$\quad f(x,y)= \dfrac{x^2y}{x^3 + y^3}$
Ta có:
$+)\quad \dfrac{\partial f}{\partial x}= \dfrac{y(2xy^3 - x^4)}{(x^3 + y^3)^2}$
$+)\quad \dfrac{\partial f}{\partial y}= \dfrac{x^2(x^3 - 2y^3)}{(x^3 + y^3)^2}$
$\Rightarrow df = \dfrac{y(2xy^3 - x^4)}{(x^3 + y^3)^2}dx + \dfrac{x^2(x^3 - 2y^3)}{(x^3 + y^3)^2}dy$
Tại điểm $(1;1)$ ta được:
$\quad df = \dfrac14dx - \dfrac14dy$
