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I heard of the result that the derivative of an elementary function is also elementary long ago. Now I want to prove it rigorously. I found this answer(I didn't comment because it was an old post):https://math.stackexchange.com/q/2195559. It actually proved that: (1). the derivative of $\exp,\log,\operatorname{id}$ and constant is elementary. (2). If we assume that f and g have elementary derivatives, then $f+g,f-g,fg,f/g,f\circ g$ also have elementary derivatives. So far so good. But what confuses me is how does this 2 facts imply the original proposition(the derivative of any elementary function is also elementary)? I mean isn't the fact (2) seems too weak?
I think the keypoint lies in the construction of the elementary functions. I use the definition of the elementary functions to be:

  1. $\exp,\log,\operatorname{id}$ and constant is elementary.
  2. the sum, difference, product, quotient, composition of 2 elementary functions is elementary.

    Need help!
cxh007
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    You should start with the definition of elementary function. They are those and only those functions that can be produced from polynomials, rational functions, trigonometric and exponential functions and their inverses by algebraic operations and composition. Since the derivatives of the listed base functions are elementary, and sum, product, quotient and chain rules transform elementary functions into elementary functions, the derivatives of all elementary functions are elementary by induction on operations used to produce them. – Conifold Feb 20 '21 at 08:09
  • I think the idea to use induction is surprising. – cxh007 Feb 20 '21 at 08:22
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    Why? When some class of objects is defined inductively using induction to prove something for all of them is the first thing that comes to mind. – Conifold Feb 20 '21 at 08:24
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    I think the idea to use induction is surprising --- Of possible interest is the general notion of inductive definitions, which can be worked with in two main ways, "from above" and "from below". A fairly abstract treatment of this notion is given in Elementary Induction on Abstract Structures by Moschovakis (1974; reprinted by Dover in 2008). – Dave L. Renfro Feb 20 '21 at 09:05
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    Yes, this is a more general type of induction/recursion than the elementary one on natural numbers, it is sometimes called structural induction. It is a very useful tool in more advanced problems. If the elementary type can be visualized as moving up a chain or a ladder, structural induction moves up a tree that can branch. In this case, you have choices which operations to apply to produce new elementary functions from old ones, so the steps can not be indexed by natural numbers like a sequence. – Conifold Feb 20 '21 at 09:09
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    Yes, this would correspond to collecting all branch points of the tree that are at the same height into a single set, but I find it more intuitive to think in terms of the tree directly. And the base set needs to also include trig functions and their inverses unless you are working over complex numbers. – Conifold Feb 20 '21 at 09:34
  • So let $E_0$ denotes the set of functions that is polynomials, rational functions, trigonometric and exponential functions and their inverses. And for every $n \in \mathbb{N}$, $n>0$, let $E_n$ denotes the set of functions which is the sum/difference/product/quotient/composition of any 2 elements(if definable) in $E_{n-1}$. Then we can use induction on this natural number $n$. – cxh007 Feb 20 '21 at 09:46

2 Answers2

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The closure under differentiation can be proven by induction.

The result is clear for « base functions » ($\ln x, e^x)...$

If a map is a sum, the product, the composition... of two simple functions, then it follows that its derivative is a simple function based on the induction hypothesis and the differentiation rules of the sum, the product, the composition... of two functions.

As any simple function can be obtained by induction using previous rules according to the very definition of a simple function, we’re done.

mathcounterexamples.net
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Let me try to organize what you have a little more neatly, maybe this will clear things up.

First, your definition of elementary function:

$ \bullet \exp, \log, \text{id}, \text{constant}\\ \bullet \text{The sum of two elementary functions, i.e. }f+g \\ \bullet \text{The difference of two elementary functions, i.e. }f-g \\ \bullet \text{The product of two elementary functions, i.e. }fg \\ \bullet \text{The quotient of two elementary functions, i.e. }\frac{f}{g} \\ \bullet \text{The composition of two elementary functions, i.e. }f \circ g $

Now, the proof you posted proves first that $\exp, \log$, id and constant functions have elementary derivatives, so the claim holds for the first bullet point in the list above. Then it goes on to prove that it holds for sums (2nd bullet point), differences (3rd bullet), products (4th bullet), quotients (5th bullet) and compositions (6th bullet).

So it really has shown that all the functions that fall under the definition of elementary function have elementary derivatives! And as mathcounterexamples.net pointed out in his answer, this can be expanded to the sum, product, etc. of more than two functions by induction.

noam.szyfer
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    I totally agree that the claim holds for the first bullet point. But when it comes to the second point, it assumes f and g both have elementary derivatives which I think can't be taken as "the claim holds for sums". – cxh007 Feb 20 '21 at 08:48
  • Ah ok I think I see what you mean. In that case Induction is really the way to go (as seen in the other answer) – noam.szyfer Feb 20 '21 at 08:54