Question

z=cos (2x^2-y) найти dz

Answer 1

$z=cos(2x^2-y)\\dz=z'_{x}dx+z'_{y}dy\\dz=-sin(2x^2-y)cdot 4xcdot dx-sin(2x^2-y)cdot (-1)cdot dy\\dz=-4xcdot sin(2x^2-y)dx+sin(2x^2-y)dy\\dz=sin(2x^2-y)cdot (-4x, dx+dy)$