Question

Найти полный дефиринциал первого и дефиринциал второго порядка
$z=\frac{x^2*y^2}{x+y} +ln(x+y)$

Буду благодарен если до пятницы.

Answer 1

$z=\dfrac{x^2\cdot y^2}{x+y}+ln(x+y)\\\\\\\dfrac{\partial z}{\partial x}}=\dfrac{2xy^2(x+y)-x^2y^2}{(x+y)^2}+\dfrac{1}{x+y}=\dfrac{2x^2y^2+2xy^3-x^2y^2}{(x+y)^2}+\dfrac{1}{x+y}=\\\\\\=\dfrac{x^2y^2+2xy^3}{(x+y)^2}+\dfrac{1}{x+y}\\\\\\\dfrac{\partial z}{\partial y}=\dfrac{2yx^2(x+y)-x^2y^2}{(x+y)^2}+\dfrac{1}{x+y}=\dfrac{2x^3y+2x^2y^2-x^2y^2}{(x+y)^2}+\dfrac{1}{x+y}=\\\\\\=\dfrac{x^2y^2+2x^3y}{(x+y)^2}+\dfrac{1}{x+y}\\\\\\dz=\dfrac{\partial z}{\partial x}\cdot dx+\dfrac{\partial z}{\partial y}\cdot dy$

$dz=\Big(\dfrac{x^2y^2+2xy^3}{(x+y)^2}+\dfrac{1}{x+y}\Big)\cdot dx+\Big(\dfrac{x^2y^2+2x^3y}{(x+y)^2}+\dfrac{1}{x+y}\Big)\cdot dy$

$\dfrac{\partial^2z}{\partial x^2} =\dfrac{(2xy^2+2y^3)(x+y)^2-2(x+y)(x^2y^2+2xy^3)}{(x+y)^4}-\dfrac{1}{(x+y)^2} =\\\\\\=\dfrac{2x^2+4xy^3+2y^4-2x^2y^2-4xy^3}{(x+y)^3}-\dfrac{1}{(x+y)^2} =\dfrac{2y^4}{(x+y)^3}-\dfrac{1}{(x+y)^2}=\\\\\\=\dfrac{2y^4-x-y}{(x+y)^3}$

$\dfrac{\partial ^2z}{\partial y^2}=\dfrac{(2x^2y+2x^3)(x+y)^2-2(x+y)(x^2y^2+2x^3y)}{(x+y)^4}-\dfrac{1}{(x+y)^2}=\\\\\\=\dfrac{4x^3y+2x^2y^2+2x^4-2x^2y^2-4x^3y}{(x+y)^3}-\dfrac{1}{(x+y)^2}=\dfrac{2x^4}{(x+y)^3}-\dfrac{1}{(x+y)^2}=\\\\\\=\dfrac{2x^4-x-y}{(x+y)^3}$

$\dfrac{\partial^2z}{\partial x\partial y}=\dfrac{(2x^2y+6xy^2)(x+y)^2-2(x+y)(x^2y^2+2xy^3)}{(x+y)^4}-\dfrac{1}{(x+y)^2}=\\\\\\=\dfrac{2x^3y+8x^2y^2+6xy^3-2x^2y^2-4xy^3}{(x+y)^3}-\dfrac{1}{(x+y)^2}=\\\\\\=\dfrac{2x^3y+6x^2y^2+2xy^3}{(x+y)^3}-\dfrac{1}{(x+y)^2}=\dfrac{2xy(x^2+2xy+y^2)-x-y}{(x+y)^3}$

$d^2z=\dfrac{\partial ^2z}{\partial x^2}\cdot dx^2+2\dfrac{\partial ^2z}{\partial x\partial y}\codt dx\, dy+\dfrac{\partial ^2z}{\partial y^2}\cdot dy^2\\\\\\d^2z=\dfrac{2y^4-x-y}{(x+y)^3}\cdot dx^2+\dfrac{4xy(x^2+3xy+y^2)-2x-2y}{(x+y)^3}\cdot dx\, dy+\dfrac{2x^4-x-y}{(x+y)^3}\cdot dy^2$