What's the exterior derivative of the two form: $x \,dy \wedge dz + y \,dx \wedge dz + z\,dx \wedge dy$?

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What's the exterior derivative of the two form: $x \,dy \wedge dz + y\, dx \wedge dz + z\,dx \wedge dy$?

I saw some calculations on this site and I'm pretty sure that the answer is $3\, dx\wedge dy\wedge dz$ , however, this conflicts with the answer posted on here, Exterior derivative of a 2-form, which I get $ dx\wedge dy\wedge dz$.

How do I do this correctly? Thanks in advance!

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Well, you just use that $d(\alpha\wedge \beta) = d\alpha\wedge \beta + \alpha\wedge d\beta$, and in particular $d(fdg) = df\wedge dg$ since $ddg = 0$.

So $d(xdy\wedge dz) = d(xdy)\wedge dz + (xdy)\wedge ddz = dx\wedge dy\wedge dz$.

Likewise, $d(ydx\wedge dz) = dy\wedge dx\wedge dz$ and $d(zdx\wedge dy) = dz\wedge dx\wedge dy$.

Now $dy\wedge dx\wedge dz = -dx\wedge dy\wedge dz$ and $dz\wedge dx\wedge dy = dx\wedge dy\wedge dz$.

All in all the sum is $dx\wedge dy\wedge dz$. You most likely forgot the signs.