# I'm calculating exterior derivatives and my answer is off by a factor of -1. What am I doing wrong?

For example. I want to find the exterior derivative of

$$\omega=(-3x-2y-z)dx +(2x^2+3y^3+4z^4)dy + (x+2y+3z)dz$$

Here's what I did:

\begin{align} d\omega &= \left[\frac{\partial}{\partial x}(-3x-2y-z)\ dx+\frac{\partial}{\partial x}(2x^2+3y^3+4z^4)\ dy+\frac{\partial}{\partial x}(x+2y+3z)\ dz\right]dx \\ &+ \left[\frac{\partial}{\partial y}(-3x-2y-z)\ dx+\frac{\partial}{\partial y}(2x^2+3y^3+4z^4)\ dy+\frac{\partial}{\partial y}(x+2y+3z)\ dz\right]dy \\ &+ \left[\frac{\partial}{\partial z}(-3x-2y-z)\ dx+\frac{\partial}{\partial z}(2x^2+3y^3+4z^4)\ dy+\frac{\partial}{\partial z}(x+2y+3z)\ dz\right]dz \\ &= (-3\ dx +4x\ dy +1\ dz)\ dx +(-2\ dx+9y^2\ dy+2\ dz)\ dy+(-1\ dx+16z^3\ dy+3\ dz)\ dz \\ &= 4x\ dy\ dx+ dz\ dx-2\ dx\ dy+2\ dz\ dy-1\ dx\ dz+16z^3\ dy\ dz \\ &= -4x\ dx\ dy-2\ dx\ dy-1\ dx\ dz-1\ dx\ dz-2\ dy\ dz+16z^3\ dy\ dz \\ &= (-4x-2)\ dx\ dy-2\ dx\ dz+(16z^3-2)\ dy\ dz \end{align}

However, Maple says this:

Maple says I'm off by a factor of -1? What am I doing wrong?

Thank you!

• the derivative of $2x^2$ is $4x\,dx$, not $-2\,dx.$ The derivative of $4z^4$ is $16z^3\,dz,$ not $2\,dz.$ – ziggurism Nov 20 '17 at 0:24
• The derivative of $x$ is $+1$, not $-1$ times $dx.$ The derivative of $2y$ is $2\,dy,$ not $16z^3\,dy$ (???) – ziggurism Nov 20 '17 at 0:27
• Maybe I'm doing things completely wrong, but I edited in a step with partial derivatives to show what I was doing. – WA1NGRO Nov 20 '17 at 0:42
• Still don't see how you go from $2x^2$ to $-2\,dx.$ – ziggurism Nov 20 '17 at 1:00
• @ziggurism $2x^2\ dy$ becomes $4x\ dy\ dx$ in the first term of OP's answer – Dylan Nov 20 '17 at 1:03

Your order of operations is wrong. If $$\omega = u\ dx + v\ dy + w\ dz$$
then \begin{align} d\omega &= \frac{\partial u}{\partial x} \ dx\wedge dx + \frac{\partial v}{\ \partial x} \ dx \wedge dy + \frac{\partial w}{\partial x}\ dx \wedge dz \\ &\ + \frac{\partial u}{\partial y} \ dy\wedge dx + \frac{\partial v}{\partial y} \ dy \wedge dy + \frac{\partial w}{\partial y}\ dy \wedge dz \\ &\ + \frac{\partial u}{\partial z} \ dz\wedge dx + \frac{\partial v}{\partial z} \ dz \wedge dy + \frac{\partial w}{\partial z}\ dz \wedge dz \end{align}
You had it written as $$d\omega = \left(\frac{\partial u}{\partial x} \ dx + \frac{\partial v}{\partial x} \ dy + \frac{\partial w}{\partial x} \ dz \right) \wedge dx + \dots$$