I need to show that if
$\omega=xydx+3dy-yzdz$
$\nu=xdx-yz^2dy+2xdz$
then $d(\omega\land\nu)=(d\omega)\land\nu-\omega\land(d\nu)$
I really do not understand how this differential operators work. I would be thankful of someone helped me with this
Thanks in advance!
On the left hand side:
$\omega\land\nu=-xy^2z^2dx\land dy + 2x^2y dx \land dz + 3x dy \land dx + 6x dy \land dz - xyz dz \land dx + y^2z^3 dz \land dy $
$ =-xy^2z^2dx\land dy + 2x^2y dx \land dz - 3x dx \land dy + 6x dy \land dz + xyz dx \land dz - y^2z^3 dy \land dz $
$ =(-xy^2z^2-3x)dx\land dy + (2x^2y+ xyz) dx \land dz + (6x- y^2z^3) dy \land dz $
$d(\omega\land\nu)=-2xy^2z dz \land dx \land dy + (2x^2+ xz) dy \land dx \land dz + 6 dx \land dy \land dz $
$ = (-2x^2 -2xy^2z -xz +6) dx \land dy \land dz $
On the right hand side:
$d\omega = x dy \land dx - z dy \land dz $
$d\nu = - 2yz dz \land dy + 2 dx \land dz $
$d\omega \land \nu = -2x^2 dx \land dy \land dx - xz dy \land dz \land dx $
$\omega \land d\nu = 2xy^2z dx \land dy \land dz + 6 dy \land dx \land dz $
$d\omega \land \nu - \omega \land d\nu = (-2xy^2z-2x^2) dx \land dy \land dz + (-6+xz) dy \land dx \land dz $
$ = (-2x^2 -2xy^2z -xz +6) dx \land dy \land dz $