Question

найти частное производное и доказать что функция
$z=\frac{x^2*y^2}{x+y}$
устраивает соотношение $xz''_{xx} +yz''_{yy} =2z_{x}$

Answer 1

$z=\dfrac{x^2\cdot y^2}{x+y}\\\\\\\dfrac{\partial z}{\partial x}=\dfrac{2xy^2(x+y)-x^2y^2}{(x+y)^2}=\dfrac{x^2y^2+2xy^3}{(x+y)^2}\\\\\\\dfrac{\partial z}{\partial y}=\dfrac{2yx^2(x+y)-x^2y^2}{(x+y)^2}=\dfrac{x^2y^2+2x^3y}{(x+y)^2}\\\\\\\dfrac{\partial^2z}{\partial x^2} =\dfrac{(2xy^2+2y^3)(x+y)^2-2(x+y)(x^2y^2+2xy^3)}{(x+y)^4}=\\\\\\=\dfrac{2x^2+4xy^3+2y^4-2x^2y^2-4xy^3}{(x+y)^3}=\dfrac{2y^4}{(x+y)^3}$

$\dfrac{\partial ^2z}{\partial y^2}=\dfrac{(2x^2y+2x^3)(x+y)^2-2(x+y)(x^2y^2+2x^3y)}{(x+y)^4}=\\\\\\=\dfrac{4x^3y+2x^2y^2+2x^4-2x^2y^2-4x^3y}{(x+y)^3}=\dfrac{2x^4}{(x+y)^3}$

$\dfrac{\partial^2z}{\partial x\partial y}=\dfrac{(2x^2y+6xy^2)(x+y)^2-2(x+y)(x^2y^2+2xy^3)}{(x+y)^4}=\\\\\\=\dfrac{2x^3y+8x^2y^2+6xy^3-2x^2y^2-4xy^3}{(x+y)^3}=\dfrac{2x^3y+6x^2y^2+2xy^3}{(x+y)^3}$

$\dfrac{\partial ^2z}{\partial x\partial y}=\dfrac{\partial ^2z}{\partial y\partial x}$

$x\, z''_{xx}+y\, z''_{yy}=\dfrac{2xy^4}{(x+y)^3}+\dfrac{2x^4y}{(x+y)^3}=\dfrac{2xy\, (y^3+x^3)}{(x+y)^3}=\\\\\\=\dfrac{2xy\, (x+y)(x^2-xy+y^2)}{(x+y)^3}=\dfrac{2xy(x^2-xy+y^2)}{(x+y)^2}$