Đáp án:
\(\begin{array}{l}
a)\\
\bullet\quad \dfrac{\partial f}{\partial x} = e^{y^2 - xy}(-2x^2y +4x- y)\\
\bullet\quad \dfrac{\partial f}{\partial y} =e^{y^2 - xy}(2y - x)(2x^2 + 1)\\
b)\\
\bullet\quad f_{x^2}''(1;0) = 4\\
\bullet\quad f_{y^2}''(1;0) = 3
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
\quad f(x,y) = e^{y^2-xy}(2x^2 + 1)\\
a)\\
\bullet\quad \dfrac{\partial f}{\partial x} = \dfrac{\partial }{\partial x}(e^{y^2 - xy})\cdot (2x^2 + 1) + \dfrac{\partial }{\partial x}(2x^2 + 1)\cdot e^{y^2 - xy}\\
\Leftrightarrow \dfrac{\partial f}{\partial x} = -y.e^{y^2 - xy}(2x^2 + 1) + 4x.e^{y^2 - xy}\\
\Leftrightarrow \dfrac{\partial f}{\partial x} = e^{y^2 - xy}(-2x^2y +4x- y)\\
\bullet\quad \dfrac{\partial f}{\partial y} = \dfrac{\partial }{\partial y}(e^{y^2 - xy})\cdot (2x^2 + 1) + \dfrac{\partial }{\partial y}(2x^2 + 1)\cdot e^{y^2 - xy}\\
\Leftrightarrow \dfrac{\partial f}{\partial y} =(2y - x)e^{y^2 - xy}(2x^2 + 1) + 0\cdot e^{y^2 - xy}\\
\Leftrightarrow \dfrac{\partial f}{\partial y} =e^{y^2 - xy}(2y - x)(2x^2 + 1)\\
b)\\
\bullet\quad f_{x^2}'' = \dfrac{\partial^2f}{\partial x^2} = \dfrac{\partial}{\partial x}\left(\dfrac{\partial f}{\partial x}\right)\\
\Leftrightarrow f_{x^2}'' = \dfrac{\partial}{\partial x}\left[e^{y^2 - xy}(-2x^2y +4x- y)\right]\\
\Leftrightarrow f_{x^2}'' =\dfrac{\partial}{\partial x}(e^{y^2 - xy})\cdot (-2x^2y +4x- y) + \dfrac{\partial}{\partial x}(-2x^2y +4x- y)\cdot e^{y^2 - xy}\\
\Leftrightarrow f_{x^2}'' =-y.e^{x^2 - xy}(-2x^2y +4x- y) + (-4xy+4).e^{y^2 - xy}\\
\Leftrightarrow f_{x^2}'' = e^{y^2 - xy}(2x^2y^2 -8xy + y^2 + 4)\\
\bullet\quad f_{y^2}'' = \dfrac{\partial^2f}{\partial y^2} = \dfrac{\partial}{\partial y}\left(\dfrac{\partial f}{\partial y}\right)\\
\Leftrightarrow f_{y^2}''=\dfrac{\partial}{\partial y}\left[e^{y^2 - xy}(2y - x)(2x^2 + 1)\right]\\
\Leftrightarrow f_{y^2}''=\dfrac{\partial}{\partial y}(e^{y^2 - xy})\cdot (2y-x)(2x^2 + 1)+ \dfrac{\partial}{\partial y}(2y-x)\cdot e^{y^2 - xy}(2x^2 + 1)\\
\Leftrightarrow f_{y^2}''=(2y - x)^2e^{y^2 - xy}(2x^2 + 1) + 2e^{y^2 - xy}\\
\Leftrightarrow f_{y^2}''=e^{y^2 - xy}(x^2 - 4xy + y^2 + 2)\\
\text{Ta được:}\\
\bullet\quad f_{x^2}''(1;0) = e^{0^2 - 1.0}(2.1^2.0^2-8.1.0+0^2 +4) = 4\\
\bullet\quad f_{y^2}''(1;0) = e^{0^2 - 1.0}(1^2 - 4.1.0 + 0^2 + 2) = 3
\end{array}\)
