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$\begingroup$

What do the square bracket symbols mean? Are they what I hear are "sets"? And when it is in an equation, how is it interpreted?

Here is an example:

$$\dfrac{dy}{dx}[2x^{2}+y(x)^{2}]=50x+2y(dy/dx)=0$$

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    It very much depends on the context. Could you show us a specific instance?2012-02-02
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    I found this on http://www.sosmath.com/calculus/diff/der05/der05.html: $$dy/dx[2x^2 + y(x)^2] = 50x + 2y(dy/dx) = 0$$2012-02-02
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    That helps out quite a bit. I think the $[\;]$ there are just parentheses by another "name"; you can replace them with $(\;)$ without fear. People will sometimes use $[\;]$ or $\{\;\}$ for this purpose in order to avoid a forest of $)))))$s. It doesn't seem so necessary here.2012-02-02
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    The square brackets. Next to dy/dx2012-02-02
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    David, since you seem to have some programming background: almost all symbols in math are heavily 'overloaded'. In your example, the squared brackets, they are simply are replacement for the are normal brackets (do you know what 'dx/dx' means? Just checking...) But they can for instance denote a closed interval as in $x\in[0,1]$, or something completely different...2012-02-02

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Sometimes they are the same as parentheses. Like in $\sum \left[ f(n) - g(n) \right]$ or the expected value $\text{E} \left[ X \right]$. They may have different meanings if they are specially defined.

Or if you meant $\{ \}$ then these are braces and yes they are often used to represent sets.

There is a lot of ambiguity in the proper nomenclature.

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    No, I don't mean those {}, I mean these []2012-02-02
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    Show us a context we'll give you an answer. With no context, those could mean anything.2012-02-02
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    See my edit ^^^2012-02-02