$f(x,y) = e^{x^2 - xy}(2x^2+1)$

a) Đạo hàm riêng cấp $1$

$\bullet\quad \dfrac{\partial f}{\partial x} = e^{x^2 - xy}(4x^3 - 2x^2y+ 6x - y)$

$\bullet\quad \dfrac{\partial f}{\partial y} = -x(2x^2+1)e^{x^2 - xy}$

b) Đạo hàm riêng cấp $2$

$\bullet\quad f_{x^2}''(x,y)= \dfrac{\partial^2f}{\partial x^2}$

$\Leftrightarrow f_{x^2}''(x,y)= \dfrac{\partial}{\partial x}\left(\dfrac{\partial f}{\partial x}\right)$

$\Leftrightarrow f_{x^2}''(x,y)=\dfrac{\partial}{\partial x}\left[e^{x^2 - xy}(4x^3 - 2x^2y+ 6x - y)\right]$

$\Leftrightarrow f_{x^2}''(x,y)=e^{x^2 - xy}[8x^4 - 8x^3y+ 2x^2(y^2 +12) -12xy + y^2 +6]$

$\bullet\quad f_{y^2}''(x,y)= \dfrac{\partial^2f}{\partial y^2}$

$\Leftrightarrow f_{y^2}''(x,y)= \dfrac{\partial}{\partial y}\left(\dfrac{\partial f}{\partial y}\right)$

$\Leftrightarrow f_{y^2}''(x,y)=\dfrac{\partial}{\partial y}\left[-x(2x^2+1)e^{x^2 - xy}\right]$

$\Leftrightarrow f_{y^2}''(x,y)= x^2(2x^2+1)e^{x^2 - xy}$

Ta được:

$\bullet\quad f_{x^2}''(0;a) = a^2 + 6$

$\bullet\quad f_{y^2}''(0;a) = 0$