Question

Найти полный дефиринциал первого и дефиринциал второго порядка
$z=arctg(\frac{x}{y} )+arcsin(\frac{x-y}{x+y} )$
Нужно сегодня пожалуйста.

Answer 1

$z=arctg\dfrac{x}{y}+arcsin\dfrac{x-y}{x+y}\\\\\\z'_{x}=\dfrac{1}{1+\frac{x^2}{y^2}}\cdot \dfrac{1}{y}+\dfrac{1}{\sqrt{1-\frac{(x-y)^2}{(x+y)^2}}}\cdot \dfrac{(x+y)-(x-y)}{(x+y)^2}=\\\\\\=\dfrac{y^2}{x^2+y^2}\cdot \dfrac{1}{y}+\dfrac{x+y}{\sqrt{4xy}}\cdot \dfrac{2y}{(x+y)^2}=\dfrac{y}{x^2+y^2}+\dfrac{\sqrt{y}}{\sqrt{x}\, (x+y)}$

$z'_{y}=\dfrac{1}{1+\frac{x^2}{y^2}}\cdot \dfrac{-x}{y^2}+\dfrac{1}{\sqrt{1-\frac{(x-y)^2}{(x+y)^2}}}\cdot \dfrac{-(x+y)-(x-y)}{(x+y)^2}=\\\\\\=\dfrac{-x}{x^2+y^2}+\dfrac{x+y}{\sqrt{4xy}}\cdot \dfrac{-2x}{(x+y)^2}=-\dfrac{x}{x^2+y^2}-\dfrac{\sqrt{x}}{\sqrt{y}\, (x+y)}$

$dz=\Big(\dfrac{y}{x^2+y^2}+\dfrac{\sqrt{y}}{\sqrt{x}\, (x+y)}\Big)\, dx+\Big(-\dfrac{x}{x^2+y^2}-\dfrac{\sqrt{x}}{\sqrt{y}\, (x+y)}\Big)\, dy$

$z''_{xx}=\dfrac{-y\cdot 2x}{(x^2+y^2)^2}+\dfrac{-\sqrt{y}\cdot (\frac{1}{2\sqrt{x}}(x+y)+\sqrt{x})}{x\cdot (x+y)^2}=\\\\\\=-\dfrac{2xy}{(x^2+y^2)^2}-\dfrac{\frac{1}{2}\sqrt{xy}+\frac{y\sqrt{y}}{2\sqrt{x}}+\sqrt{xy}}{x\cdot (x+y)^2}=-\dfrac{2xy}{(x^2+y^2)^2}-\dfrac{x\sqrt{y}+y\sqrt{y}+2x\sqrt{y}}{2\sqrt{x}\cdot x\cdot (x+y)^2}=\\\\\\=-\dfrac{2xy}{(x^2+y^2)^2}-\dfrac{\sqrt{y}\cdot (3x+y)}{2\sqrt{x}\cdot x(x+y)^2}$

$z''_{yy}=\dfrac{x\cdot 2y}{(x^2+y^2)^2}+\dfrac{\sqrt{x}\cdot (\frac{1}{2\sqrt{y}}(x+y)+\sqrt{y})}{y(x+y)^2}=\\\\\\=\dfrac{2xy}{(x^2+y^2)^2}+\dfrac{\frac{x\sqrt{x}}{2\sqrt{y}}+\frac{1}{2}\sqrt{xy}+\sqrt{xy}}{y(x+y)^2}=\dfrac{2xy}{(x^2+y^2)^2}+\dfrac{\frac{x\sqrt{x}}{2\sqrt{y}}+\frac{3}{2}\sqrt{xy}}{y(x+y)^2}=\\\\\\=\dfrac{2xy}{(x^2+y^2)^2}+\dfrac{x\sqrt{x}+3y\sqrt{x}}{2\sqrt{y}\cdot y(x+y)^2}=\dfrac{2xy}{(x^2+y^2)^2}+\dfrac{\sqrt{x}(x+3y)}{2\sqrt{y}\cdot y(x+y)^2}$

$z''_{xy}=\dfrac{x^2+y^2-y\cdot 2y}{(x^2+y^2)^2}+\dfrac{\frac{1}{2\sqrt{y}}\cdot \sqrt{x}(x+y)-\sqrt{y}\cdot \sqrt{x}}{x(x+y)^2}=\\\\\\=\dfrac{x^2-y^2}{(x^2+y^2)^2}+\dfrac{\frac{x\sqrt{x}}{2\sqrt{y}}+\frac{1}{2}\sqrt{xy}-\sqrt{xy}}{x(x+y)^2}=\dfrac{x^2-y^2}{(x^2+y^2)^2}+\dfrac{\frac{x\sqrt{x}}{2\sqrt{y}}-\frac{1}{2}\sqrt{xy}}{x(x+y)^2}=\\\\\\=\dfrac{x^2-y^2}{(x^2+y^2)^2}+\dfrac{x\sqrt{x}-y\sqrt{x}}{2\sqrt{y}\cdot x(x+y)^2}=\dfrac{x^2-y^2}{(x^2+y^2)^2}+\dfrac{x-y}{2\sqrt{xy}(x+y)^2}$

$d^2z=\Big(-\dfrac{2xy}{(x^2+y^2)^2}-\dfrac{\sqrt{y}\, (3x+y)}{2\, x\sqrt{x}\, (x+y)^2}\Big)\, dx^2+\Big(\dfrac{x^2-y^2}{(x^2+y^2)^2}+\dfrac{x-y}{2\sqrt{xy}(x+y)^2}\Big)\, dx\, dy+\\\\\\+\Big(\dfrac{2xy}{(x^2+y^2)^2}+\dfrac{\sqrt{x}\, (x+3y)}{2\, y\sqrt{y}\, (x+y)^2} \Big)\, dy^2$