Помогите задали решить 2 примера. 1 по теме дифференциал функций двух переменных. Другое...

Решите задачу:

$1) \ z=arctg\frac{x-2}{y}, \\ \frac{\partial z}{\partial x} = (arctg\frac{x-2}{y})'_x = \frac{1}{1+(\frac{x-2}{y})^2}\cdot(\frac{x-2}{y})'_x = \frac{y^2}{y^2+(x-2)^2}\cdot\frac{1}{y} = \frac{y}{y^2+(x-2)^2}; \\ \frac{\partial z}{\partial y} = (arctg\frac{x-2}{y})'_y = \frac{1}{1+(\frac{x-2}{y})^2}\cdot(\frac{x-2}{y})'_y = \frac{y^2}{y^2+(x-2)^2}\cdot(-\frac{x-2}{y^2}) =\\= \frac{2-x}{y^2+(x-2)^2},$
$(x-2)\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y} = (x-2)\cdot\frac{y}{y^2+(x-2)^2}+y\cdot\frac{2-x}{y^2+(x-2)^2} =\\= \frac{y(x-2)}{y^2+(x-2)^2} - \frac{y(x-2)}{y^2+(x-2)^2} = 0;$

$z=x^2-4x+y^2, \\ z'_x=2x-4, \\ z'_y=2y, \\ z''_{xx}=2, \ z''_{xy}=0, \\ z''_{yx}=0, \ z''_{yy}=2. \\ z''_{xy}=z''_{yx}, \\ \left \{ {{z'_x=0,} \atop {z'_y=0;}} \right. \ \left \{ {{2x-4=0,} \atop {2y=0;}} \right. \ \left \{ {{x=2,} \atop {y=0;}} \right. \\ M(2;0). \\ A=z''_{xx}(M)=2\ \textgreater \ 0, \ B=z''_{xy}(M)=0, \ C=z''_{yy}(M)=2, \\ AC-B^2=2\cdot2-0=4\ \textgreater \ 0, \\ M(2;0)-min.$