Question

докажите тождество если $z= arctg(frac{x}{x^2+y^2} )$ то $frac{d^2z}{dxdy}=frac{d^2z}{dydx}$

Answer 1

$frac{dz}{dx} = frac{1}{1 +(frac{x}{x^2 + y^2})^2} cdot frac{x^2 + y^2 - 2x^2}{(x^2 + y^2)^2} = frac{y^2 - x^2}{x^4 + 2x^2y^2 + y^4 + x^2}$.

$frac{dz}{dy} = frac{1}{1 + (frac{x}{x^2 + y^2})^2} cdot (-frac{2xy}{(x^2 + y^2)^2}) = -frac{2xy}{x^4 + 2x^2y^2 + y^4 + x^2}$.

$frac{d^2 z}{dxdy} = frac{2y(x^4 + 2x^2y^2 + y^4 + x^2) - (y^2 - x^2)(4yx^2 + 4y^3)}{(x^4 + 2x^2y^2 + y^4 + x^2)^2}=\\ = frac{2 y (3 x^4 - y^4 + 2x^2y^2 + x^2)}{(x^4 + y^4 + 2x^2y^2 + x^2)^2}$.

$frac{d^2z}{dydx} = frac{(4x^3 + 4xy^2 + 2x)2xy - 2x(x^4 + 2x^2y^2 + y^4 + x^2)}{(x^4 + 2x^2y^2 + y^4 + x^2)^2} = \\= frac{2 y (3 x^4 - y^4 + 2x^2y^2 + x^2)}{(x^4 + y^4 + 2x^2y^2 + x^2)^2}$.

Откуда $frac{d^2 z}{dxdy} = frac{d^2 z}{dydx}$.