Question

Найти частные производыне функции $z=arcsin(\frac{x-y}{x+y} )$

Answer 1

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

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

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

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

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