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UwU brat mathematician behavior (lemmynsfw.com)

submitted 5 days ago by [email protected] to c/[email protected]

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[–] [email protected] 2 points 4 days ago (1 children)

The operand is the target of an operator

Correct. Thus, dx is an operand. It's a thing by which you multiply the rest of the equation (or, in the case of dy/dx, by which you divide the dy).

[–] [email protected] 2 points 4 days ago (1 children)

I'd say the $\int dx$ is the operator and the integrand is the operand.

[–] [email protected] 1 points 4 days ago (1 children)

You're misunderstanding the post. Yes, the reality of maths is that the integral is an operator. But the post talks about how "dx can be treated as an [operand]". And this is true, in many (but not all) circumstances.

∫(dy/dx)dx = ∫dy = y

Or the chain rule:

(dz/dy)(dy/dx) = dz/dx

In both of these cases, dx or dy behave like operands, since we can "cancel" them through division. This isn't rigorous maths, but it's a frequently-useful shorthand.

[–] [email protected] 1 points 4 days ago

I do understand it differently, but I don't think I misunderstood. I think what they meant is the physicist notation I'm (as a physicist) all too familiar with:

∫ f(x) dx = ∫ dx f(x)

In this case, because f(x) is the operand and ∫ dx the operator, it's still uniquely defined.