Найти частные производные второго порядка от функций: 1) z=x*y+[latex] \frac{x}{y} [/latex] 2)z=ln([latex] x^{2} [/latex] +[latex] y^{2} [/latex])

[latex]\huge 1.\\ z=xy+\frac{x}{y}\\ \frac{\partial z}{\partial x}=y+\frac{1}{y}\\ \frac{\partial z}{\partial y}=x-\frac{x}{y^2}\\ \frac{\partial^2 z}{\partial x^2}=0\\ \frac{\partial^2 z}{\partial y^2}=\frac{2x}{y^3}\\ \frac{\partial^2 z}{\partial x \partial y}=\frac{\partial^2 z}{\partial y \partial x}=1-\frac{1}{y^2}\\ \\2.\\ z=ln(x^2+y^2)\\ \frac{\partial z}{\partial x}=\frac{1}{x^2+y^2}*2x=\frac{2x}{x^2+y^2}\\ \frac{\partial z}{\partial y}=\frac{1}{x^2+y^2}*2y=\frac{2y}{x^2+y^2}\\ \frac{\partial^2 z}{\partial x^2}=\frac{2(y^2-x^2)}{(x^2+y^2)^2}\\ \frac{\partial^2 z}{\partial y^2}=\frac{2(x^2-y^2)}{(x^2+y^2)^2}\\ \frac{\partial^2 z}{\partial x \partial y}=\frac{\partial^2 z}{\partial y \partial x}=\frac{-4xy}{(x^2+y^2)^2}\\[/latex]