Considering many elementary functions have an antiderivative which is not elementary, why does this type of thing not also happen in differential calculus?

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    $\begingroup$ Looks more of a coincidence to me. The most significant thing is that we have a load of rules for derivatives, including derivative of composition, so this boild down to some classical functions like $\exp,$\ln,\sin,\cos$, etc. Those were often obtained from some practical (mostly physical) applications, and defined through some differential equations, which give you their derivatives. $\endgroup$ – lisyarus Mar 20 '17 at 9:03
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    $\begingroup$ @lisyarus The origin in physical phenomena and the definitions from differential equations are not arguments to justify the existence of closed-forms. $\endgroup$ – Yves Daoust Mar 20 '17 at 10:45
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    $\begingroup$ If we can differentiate $f(x)$ and $g(x)$ then we can usually easily differentiate $f(x)+g(x)$, $f(x)g(x)$, $\frac{f(x)}{g(x)}$ and $f(g(x))$. $\endgroup$ – Henry Mar 20 '17 at 17:19
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    $\begingroup$ I wonder if there's a deep reason (e.g., category theory related, perhaps) $\endgroup$ – MathematicsStudent1122 Mar 20 '17 at 19:55
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    $\begingroup$ @lisyarus Antidifferentiation also has practical physical applications. $\endgroup$ – immibis Mar 20 '17 at 20:52

Just think of how we find those elementary functions:

  • We start with the constant functions, which have derivative $0$, and the identity function $f(x)=x$ which has derivative $1$.

  • We combine functions by means of addition, subtraction, multiplication, division, composition. For all of those cases we have explicit rules for the derivative.

  • We define new functions as the integral of other functions (e.g. $\ln x$ as integral of $1/x$). Obviously when deriving those we get back the function we started with.

  • We define functions as the inverse of another function. Again, we've got an explicit formula for derivatives of inverse functions.

Any function that cannot be defined by a chain of such operations (and also some which can, using the integration rule) we don't consider elementary.

So basically the reason is in the way we construct elementary functions. In some sense, one could say it is because of what functions we consider elementary.

Indeed, this hold not only for elementary functions; even most non-elementary functions we use are defined through such operations (in particular by integrals).

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    $\begingroup$ The question could be recast as "why are there differentiation rules" vs. "why are there no antidifferentiation rules". $\endgroup$ – Yves Daoust Mar 20 '17 at 10:41
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    $\begingroup$ Actually the question should not mention antiderivatives at all, since it threw off most other answers, as far as I can tell. I think the OP only mentioned antiderivatives as a contrast, not as the "meat" of the topic. In this sense, this answer seems hitting the mark very well. $\endgroup$ – AnoE Mar 20 '17 at 12:55
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    $\begingroup$ Obviously ln(x) is also an elementary function, because it is the inverse of the exponential function. I thought that any function which can be obtained without integration, and without resorting to power series can be considered elementary. $\endgroup$ – Ian Krasnow Mar 20 '17 at 22:40
  • $\begingroup$ @AnoE Yes, I understand Liouville's Theorem, I only mentioned antiderivatives for contrast. $\endgroup$ – Ian Krasnow Mar 20 '17 at 22:42
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    $\begingroup$ Weren't logarithms known before their relationship to $\frac{1}{x}$? $\endgroup$ – jpmc26 Mar 21 '17 at 1:59

The short answer is that we have differentiation rules for all the elementary functions, and we have differentiation rules for every way we can combine elementary functions (addition, multiplication, composition), where the derivative of a combination of two functions may be expressed using the functions, their derivatives and the different forms of combination.

Integration, on the other hand, neither has a direct rule for multiplication of two functions nor for composition of two functions. We can integrate the corresponding rules for differentiation and get something that looks like it (integration by parts and substitution), but it only works if you're lucky with what elementary functions are combined in what way.

You might say that there is a hope that there are rules out there, that we just haven't found them yet. This is not true; it's been proven that there are always integrals of elementary functions that are not elementary themselves (under most reasonable definitions of "elementary functions"). It's a deep result known as Liouville's theorem.

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    $\begingroup$ But why do we have rules in one case and not the other ? (Though in a way, Liouville's theorem says that the pattern of the differentiation rules prevents the existence of integration rules.) $\endgroup$ – Yves Daoust Mar 20 '17 at 10:49
  • $\begingroup$ @YvesDaoust That I do not know. $\endgroup$ – Arthur Mar 20 '17 at 10:56
  • $\begingroup$ This is what makes the question quite interesting. $\endgroup$ – Yves Daoust Mar 20 '17 at 11:20
  • $\begingroup$ It's why I called it "The short answer". $\endgroup$ – Arthur Mar 20 '17 at 11:23
  • $\begingroup$ See also this question. $\endgroup$ – Yvan Velenik Mar 20 '17 at 18:13

$\exp$, $\sin$, $\cos$, $\log$... are solutions of very simple differential equations. This fact plus the existence of explicit rules for the derivatives of sum/difference/product/quotient/composition guarantee that no "new" functions will appear when deriving elementary functions.

  • $\begingroup$ These functions are also solutions of very simple integral equations. There are also some rules for integration (not much, admittedly), but why don't they guarantee that no new function appears ? And why are the elementary functions solutions of simple differential equations ? ... $\endgroup$ – Yves Daoust Mar 20 '17 at 11:28
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    $\begingroup$ @YvesDaoust, (1) the condition is necessary but not sufficient. (2) no rule for integral of product to begin with. (3) $\exp$/$\log$ are homomorphisms ($\log$ was defined to be a homomorphism). This algebraic condition implies what will be its derivatives. A similar (geometric) reasoning can be done for $\sin$/$\cos$. And in the last pass to calculate explicitly the derivative of an elementary function you have to use that $\exp' = \exp$... $\endgroup$ – Martín-Blas Pérez Pinilla Mar 20 '17 at 11:41
  • $\begingroup$ Integration by part is a rule for the product. It expresses the integral of a product of functions in terms of another integral. The product rule for derivatives does nothing else. $\endgroup$ – Yves Daoust Mar 20 '17 at 11:47
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    $\begingroup$ @YvesDaoust, the rule of product for $fg$ gives $(fg)'$ in function of $f'$, and $g'$, not another function. If you know $f'$ an $g'$, then you know $(fg)'$ always. $\endgroup$ – Martín-Blas Pérez Pinilla Mar 20 '17 at 11:50
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    $\begingroup$ @YvesDaoust, $\int(f'\int g)$ requires a rule for the integral of a product. And for your question about homomorphism, I repeat (1). $\endgroup$ – Martín-Blas Pérez Pinilla Mar 20 '17 at 11:54

We can prove this:

From Wikipedia, an elementary function is one of the types:

  1. $\exp,\log,\operatorname{id}$
  2. Constant
  3. The sum of two elementary functions.
  4. The product of two elementary functions.
  5. The difference of two elementary functions.
  6. The quotient of two elementary functions.
  7. The composition of elementary functions.

We shall show that every case has an elementary derivative.

  1. $\mathrm D\exp = \exp$, which is of category (1). $\mathrm D\log x = x^{-1}$, which are elementary, with $x^{-1}$ being of category (6) -> (2) -> (1)
  2. The derivative of a constant is $0$, which is elementary of category (2).
  3. Follows from the linearity of differentiation: $\mathrm D(f+g) = \mathrm Df + \mathrm Dg$. Under the assumption that $f,g$ has elementary derivatives, the derivative of $f+g$ is elementary.
  4. Similar to (3), via the product rule.
  5. (2),(3),(4)
  6. Similar to (3), via the quotient rule.
  7. Chain rule.

Since every elementary function must belong to one of the categories above, it must have an elementary derivative.

  • $\begingroup$ Why does this work for derivatives and not for integrals ? $\endgroup$ – Yves Daoust Mar 21 '17 at 15:37
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    $\begingroup$ @YvesDaoust There are no rules of anti-differentiation. In particular, (4), (6), (7) fails to have elementary anti-derivatives in most cases. $\endgroup$ – Henricus V. Mar 21 '17 at 15:37
  • $\begingroup$ Of course, I know that (anyway we have the Risch algorithm). The question is why there is such an asymmetry. $\endgroup$ – Yves Daoust Mar 21 '17 at 15:49
  • $\begingroup$ @YvesDaoust For that you need to ask Liouville $\endgroup$ – Henricus V. Mar 21 '17 at 15:50
  • $\begingroup$ One can also wonder why formal differentiation works at all. For more exotic functions, such as Gamma, the derivative is out of reach. $\endgroup$ – Yves Daoust Mar 21 '17 at 15:51

Elementary functions are linear combination or quotient of polynomial and exponential functions (yes, that include sin, cos, tan). No doubt polynomials has derivative to be polynomials. And exponential is the single most important function in the universe, and clearly it is always differentiable and produces only exponential functions. The whole process of taking derivative is a combination of above process, so they are certainly elementary.

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    $\begingroup$ Does this really answer the "why" ? $\endgroup$ – Yves Daoust Mar 20 '17 at 10:39
  • for the same reason that there are strict rules for grammar, but not for creating poems. Differentiation goes ‘downwards’ – in the direction of gravity, so to speak – whereas integration goes ‘upwards’ – against the pull of gravity, so to speak. The price of defeating gravity is ambiguity, which takes the form of a lack of strict rules. Notice that integration gives you a taste of ambiguity from the very start: changing the value of a function a single point does not change the value of the integral.

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