# Differential fields and rings

If one is to compute the derivative of

$$y=3x+2$$ by $$\frac{\mathrm{d}(3x+2)}{\mathrm{d} x}$$ Would I be working with differential fields? Since differential fields is a first-order theory, but $y$ is a function of $x$ so the derivative function $d$ is a function that takes functions as arguments making it a second-order function.

I must be missing something here...

Thanks

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Not sure I understand the question. A first-order theory can speak about things that intuitively represent functions (or functions of functions and so forth), and quantify over them. Set theory does that all of the time, for example. You only need second-order logic to do that if you want to be sure that every function you can speak of at the metalevel can also be reified as an object within the theory. –  Henning Makholm Sep 7 '11 at 14:05
Right. So, say I have a theory about a shop inventory and the individuals in the universe represent items in the inventory. Even though ZF is a first-order theory, if I am to quantify over sets of these items, would I need SOL? –  kate.r Sep 7 '11 at 14:15
If you want to quantify over something in a first-order theory, then of course the "something" needs to be individuals (or possibly individuals satisfying some restriction). However, in a typical differential algebra, the individuals are already functions (the language does not allow you to speak easily about points in the domain and range of those functions). The derivation operator takes functions to functions, but is not itself something you'd need to quantify over in the theory itself -- logically it's just a single "function letter". –  Henning Makholm Sep 7 '11 at 14:54
But if I am to take the derivative of 3x+2 in terms of x, isn't 3x+2 a function and something that is not in the universe? –  kate.r Sep 7 '11 at 15:46
$3x+2$ is an element of the differential field. Thus, logically, just an individual. –  Henning Makholm Sep 7 '11 at 18:22