Question

Решить: z=(x^2+y^2)/(x-y). вычислить dz/dx+dz/dy

Answer 1

$z=\frac{x^2+y^2}{x-y}$
$\frac{dz}{dx} =(\frac{x^2+y^2}{x-y})'_x= \frac{2x(x-y)-(x^2+y^2)}{(x-y)^2} =\frac{x^2-2xy-y^2}{(x-y)^2}$

$\frac{dz}{dy} =(\frac{x^2+y^2}{x-y})'_y= \frac{2y(x-y)+(x^2+y^2)}{(x-y)^2} = \frac{2xy+x^2-y^2}{(x-y)^2}$

$\frac{dz}{dx} +\frac{dz}{dy}=\frac{x^2-2xy-y^2}{(x-y)^2}+\frac{2xy+x^2-y^2}{(x-y)^2}= \frac{2x^2-2y^2}{(x-y)^2} =\frac{2(x-y)(x+y)}{(x-y)^2}= \frac{2(x+y)}{x-y}$