Deriving spatial derivatives

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slurms
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Homework Statement
I'm trying to derive the various spatial derivatives, and find that I'm missing pieces. I'd like another set of eyes please.
Relevant Equations
div, grad, curl, grad^2
See photo. First, I see that I get more standard results by crossing theta before phi. What's the importance of that? Also I'd get more standard results by allowing some r or sine factors to leave their derivative enclosures and unify.
 

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on Phys.org
Your images are very difficult to read, due to how small the images appear. This is one of the reasons that we recommend against posting hand-written images of work.
 
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Mark44 said:
Your images are very difficult to read, due to how small the images appear. This is one of the reasons that we recommend against posting hand-written images of work
I actually figured out at least part of my answer. I was ignoring the space made by r x theta = phi, or more specifically i was working in the incorrect space r x phi = theta, which is why some of my components are the negative of the accepted answers. I figured this out by looking at my drawing, where on the upper left of my sphere the r meets the r theta and the r phi sin theta.

There's my second question tho. I derived the gradient by comparing to the results of the cross, assuming the del operator there is the gradient. But mine contain variables within the partials. I wonder if the accepted answers allow those to be factored out because they're unrelated to the associated scalae function's component
 
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haruspex said:
Which is still illegible.
you read the rest of the paragraph. This is it?
 
Mark44 said:
He's referring to the images you posted.
I know. But I did describe the problem I was having. I didn't realize how low-res rhe images would come out. Initially I posted the entire page, and then I divided it in 2 like this. My work is kinda legible enough if you know what you're looking for
 
slurms said:
But I did describe the problem I was having.
But that doesn't help much if we can't see the work.

slurms said:
My work is kinda legible enough
That's a pretty subjective assessment. So far, two people say no vs. one who says yes.
 
ok, I'll just describe what I did. First I drew a sphere and arrows representing the basis vectors: r, r theta, r phi sin theta. Where they meet, r x theta = phi. I see that maintaining that implicit order for the determinant of the curl results in the correct results for the curl: {r, theta, phi}.

I defined unit vector path and vector area, dl and da, and I note that (1/3) dl dot da = dV, the unit volume. I undid two vector integrals to create both the divergence and the curl: for vector function U, its closed integral dot da = integral divergence U dV; and also closed integral curl U dot da = closed integrals U dot dl.

For the curl, I wrote out the results of the determinant of column vectors (r, del r, U r), (theta, del theta, U theta), and so for phi. Then I followed a similar procedure as the divergence, undoing the integrals by direct comparison.

For the gradient, I assumed the dels within the column vectors of the curl were gradients. I listed the results of the curl, and then took the most general form of each del operator as the gradient.