Найдите частные производные второго порядка функции переменных[latex]u= \frac{y^2}{x+z} [/latex]

[latex]u= \dfrac{y^2}{x+z}[/latex] Частные производные первого порядка: [latex]\dfrac{\partial u }{\partial x} =\left(\dfrac{y^2}{x+z} \right)'_x= y^2\left(\dfrac{1}{x+z} \right)'_x=y^2\left(-\dfrac{1}{\left(x+z\right)^2} \right)= -\dfrac{y^2}{\left(x+z\right)^2}[/latex] [latex]\dfrac{\partial u }{\partial y} =\left(\dfrac{y^2}{x+z} \right)'_y= \dfrac{1}{x+z} \left(y^2 \right)'_y=\dfrac{1}{x+z} \cdot2y=\dfrac{2y}{x+z}[/latex] [latex]\dfrac{\partial u }{\partial z} =\left(\dfrac{y^2}{x+z} \right)'_z= y^2\left(\dfrac{1}{x+z} \right)'_z=y^2\left(-\dfrac{1}{\left(x+z\right)^2} \right)= -\dfrac{y^2}{\left(x+z\right)^2}[/latex] Заметим, что [latex] \dfrac{\partial u }{\partial x} = \dfrac{\partial u }{\partial z} [/latex] Частные производные второго порядка: [latex]\dfrac{\partial^2 u }{\partial x^2} =\left(-\dfrac{y^2}{(x+z)^2} \right)'_x= -y^2\left(\left(x+z\right)^{-2}} \right)'_x= \\\ =-y^2\cdot \left(-2\left(x+z\right)^{-3}\right)= \dfrac{2y^2}{\left(x+z\right)^3}[/latex] [latex]\dfrac{\partial^2 u }{\partial x\partial y} = \dfrac{\partial^2 u }{\partial z\partial y} =\left(-\dfrac{y^2}{(x+z)^2} \right)'_y= - \dfrac{1}{\left(x+z\right)^2} \left(y^2 \right)'_y= \\\ =- \dfrac{1}{\left(x+z\right)^2} \cdot2y=- \dfrac{2y}{\left(x+z\right)^2}[/latex] [latex]\dfrac{\partial^2 u }{\partial x\partial z} =\left(-\dfrac{y^2}{\left(x+z\right)^2} \right)'_z= -y^2\left(\left(x+z\right)^{-2}} \right)'_z= \\\ =-y^2\cdot \left(-2\left(x+z\right)^{-3}\right)= \dfrac{2y^2}{\left(x+z\right)^3} [/latex] [latex]\dfrac{\partial^2 u }{\partial y^2} =\left(\dfrac{2y}{x+z}\right)'_y= \dfrac{1}{x+z} \left(2y\right)'_y= \dfrac{1}{x+z} \cdot2= \dfrac{2}{x+z}[/latex] [latex]\dfrac{\partial^2 u }{\partial z^2} =\left( -\dfrac{y^2}{\left(x+z\right)^2} \right)'_z= -y^2\left( \left(x+z\right)^{-2} \right)'_z= \\\ =-y^2\cdot \left(-2 \left(x+z)^{-3}\right)\right)= \dfrac{2y^2}{\left(x+z\right)^3} [/latex]