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