Ответ:

Частные производные второго порядка:  [tex]\bf z''_{xx}=\dfrac{\partial ^2z}{\partial x^2}=z''_{xy}=\dfrac{\partial }{\partial x}\Big(\dfrac{\partial z}{\partial x}\Big)[/tex]  ,

[tex]\bf z''_{xy}=\dfrac{\partial ^2z}{\partial x\partial y}=\dfrac{\partial }{\partial x}\Big(\dfrac{\partial z}{\partial y}\Big)\ \ ,\ \ z''_{yx}=\dfrac{\partial ^2z}{\partial y\partial x}=\dfrac{\partial }{\partial y}\Big(\dfrac{\partial z}{\partial x}\Big)\ \ ,[/tex]  

[tex]\bf z''_{yy}=\dfrac{\partial ^2z}{\partial y^2}=\dfrac{\partial }{\partial y}\Big(\dfrac{\partial z}{\partial y}\Big)[/tex]   .  

Когда находим частную производную по одной переменной, то вторая переменная считается константой .

[tex]a)\ \ z=x^3-4x^2y+5y^2\ \ ,\\\\z'_{x}=\dfrac{\partial z}{\partial x}=3x^2-4y\cdot 2x=3x^2-8xy\\\\\\z'_{y}=\dfrac{\partial z}{\partial y}=-4x^2+10y\\\\\\z''_{xx}=\dfrac{\partial ^2z}{\partial x^2}=(3x^2-8xy)'_{x}=6x-8y\\\\\\\z''_{xy}=\dfrac{\partial ^2z}{\partial x\partial y}=(3x^2-8xy)'_{y}=-8x\ \ ,\ \ z''_{yx}=\dfrac{\partial ^2z}{\partial y\partial x}=(-4x^2+10y)'_{x}=-8x\\\\\\z''_{yy}=\dfrac{\partial ^2z}{\partial y^2}=(-4x^2+10y)'_{y}=10[/tex]  

[tex]b)\ \ z=x\cdot arctgy\ \ ,\\\\z'_{x}=\dfrac{\partial z}{\partial x}=arctgy\\\\\\z'_{y}=\dfrac{\partial z}{\partial y}=x\cdot \dfrac{1}{1+y^2}\\\\\\z''_{xx}=\dfrac{\partial ^2z}{\partial x^2}=(arctgy)'_{x}=0\\\\\\z''_{xy}=\dfrac{\partial ^2z}{\partial x\partial y}=(arctgy)'_{y}=\dfrac{1}{1+y^2}\ \ ,\ \ z''_{yx}=\dfrac{\partial ^2z}{\partial y\partial x}=\Big(x\cdot \dfrac{1}{1+y^2}\Big)'_{x}=\dfrac{1}{1+y^2}[/tex]  

[tex]z''_{yy}=\dfrac{\partial ^2z}{\partial y^2}=\Big(x\cdot \dfrac{1}{1+y^2}\Big)'_{y}=x\cdot \dfrac{-1\cdot 2y}{(1+y^2)^2}=-\dfrac{2xy}{(1+y^2)^2}[/tex]