# Digging deeper into (implicit) differentiation and integration

1. Feb 26, 2016

### NotEuler

Hello all,

Here's something I've been trying to wrap my head around:
In general, it seems that integration is 'harder' than differentiation. At least analytically. Numerically it may be the other way round.

For one thing, it's often easy to differentiate implicit functions. For example, exp(y)+y=x implicitly defines a function y(x), even though we can't solve it explicitly with elementary functions. But we can still differentiate it implicitly by differentiating both sides for x:
dy/dx exp(y)+dy/dx=1, and solving for
dy/dx=1/(exp(y)+1)

So this is an analytical solution for dy/dx, although it depends on knowledge of y to compute its value.
As a side note, would we call this an explicit solution for dy/dx? I'm not sure, since y appears on the right side (but dy/dx does not).

But what I've really been wondering about is that as far as I know, there is no way to implicitly integrate y from this equation. It seems that we would need prior knowledge of y to do this.
Is this correct, and there is no general way to integrate an implicit function?

But what is really the fundamental reason for this differences between differentiating and integrating? On some level it seems to me that in differentiating we are discarding some information, so in a sense it's easier and requires less knowledge of the mathematical relationships.
In integration, I'm not sure we can quite say the opposite ('adding' information, except in the sense of the integration constant), but we're not discarding any.
So perhaps that's one way of looking at the difference, and the reason why integration is 'harder'.
But it's not really a very satisfying and concrete answer.

I know this is a bit of a vague post, but does anyone have any thoughts on the topic?

Sincerely,
Not Euler

2. Feb 27, 2016

### Samy_A

That is a differential equation in y. As far as I know it has no solution in terms of elementary functions.
The reason that differentiation seems easier than integration is probably due to the following.
When you have a function $f$ that is defined using the well known elementary functions (trigonometric functions, polynomials, roots, exponentials, logarithms, ...), calculating the derivative is easy: use the basic rules for differentiation of sums, products, quotients, the chain rule, ...
Even for a silly function as $$f(x)=\frac{e^{\sin(\cos²(7e^x))}}{x^8+4\cos(3+\log(x^6+32))}$$
the derivative can be computed with some patience: just apply the rules.

For integration you just don't have these rules, so that even for seemingly nice functions as $$g(x)=\frac{2}{\sqrt {\pi}}e^{-x²}$$
we don't have a solution in terms of the elementary functions.
Of course, nothing forbids us to define new functions. For example, the error function is defined as the integral of the function $g$.

If this is tl;dr, then the one sentence answer is:
The derivative of an elementary function is an elementary function, the indefinite integral of an elementary function not necessarily.

Last edited: Feb 27, 2016
3. Feb 27, 2016

### NotEuler

That's really interesting. But it does make me wonder... When comparing differentiation and integration, you write "For integration you just don't have these rules...".

So what's the reason we don't, or can't have those rules? Is it because integration is defined as anti-differentiation, and it's difficult to come up with rules for such a reverse process? But I suppose we could rigorously define integration, and then define differentiation as anti-integration, so perhaps that's not a valid point either.

So does this mean that it's not possible, even in principle, to have a rule for implicit integration (i.e. an analogy of implicit differentiation)?
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4. Feb 27, 2016

### Samy_A

Implicit differentiation doesn't necessarily lead to a solution in terms of elementary functions, as your example in the first post shows.

Integration can be defined rigorously, that is done in a very general context in measure theory.
So you are correct, that integration seems more difficult than differentiation is not because it is introduced as the antiderivative in elementary calculus.

But the methods at hand for "solving" an indefinite integral are limited: basically we have to use the fundamental theorem of calculus. No algorithm at hand as we have for differentiation.
The derivative of a product $fg$ is just $fg'+f'g$. Rhetorical question: what is $\int f(x)g(x) dx$ in terms of $\int f(x) dx$ and $\int g(x) dx$?

5. Feb 28, 2016

### NotEuler

Yes, that makes sense Samy.
It does raise the question in my mind though - Why can we have no algorithm at hand for integration? The argument feels a bit circular. Is it to do with the nature of the fundamental theorem of calculus?

Or is the whole question a bit silly in the first place?

6. Feb 28, 2016

### Samy_A

It's not silly, I'm far for sure I know the definite answer to that question.

That we have an algorithm for the differentiation of elementary functions is due to the properties of differentiation seen as an operator. You can precisely describe how differentiation works on a product, on a quotient, on a composition of functions, ...
There are no such rules for integration.

It has nothing to do with the fundamental theorem of calculus. On the contrary, that theorem gives us the only "method" to solve some integrals analytically.

7. Feb 28, 2016

### HallsofIvy

Staff Emeritus
In many fields of mathematics we can distinguish between a "direct method" and an "inverse method". A "direct method" is something for which we are given a formula like "y= f(x)". Given x, we find y by applying that formula- by "doing what we are told". An "inverse method", then, would be "given the same formula and a value of y, find x". Inverse methods are typically harder than direct methods because we must first find the "inverse" formula which may not exist. Given "y= f(x)" and a value for y there may exit NO corresponding x or there may exist many such values for x.

Differentiation is a "direct method" since we are given a specific formulaic definition of the derivative, $\lim_{h\to 0}\frac{f(x+h)- f(x)}{h}$. Integration is an "indirect method" because the "anti-derivative" is defined simply as "the inverse of differentiation".

8. Feb 28, 2016

### NotEuler

Thanks HallsofIvy. Yes, I can see how the definition of the derivative is quite simple and formulaic.

But still playing devil's advocate (or just not really understanding): Is this not contradictory with what Samy wrote?
If I understand Samy correctly, we could just define integration first, and say differentiation is the inverse of integration.. But HallsofIvy says that one reason integration is more difficult is that it is defined as the inverse of differentiation.

Intuitively, my first thought was what you (HallsofIvy) wrote: that integration is hard because of it's 'inverse operation' nature, but it seemed to me that Samy's point more or less means this is not the case.

9. Feb 29, 2016

### Samy_A

Of course integration can be defined by itself. The first rigorous definition I'm aware of is the one by Riemann, called the Riemann integral.
Nowadays, for theoretical reasons, one uses the Lebesgue integral, defined in the context of measure theory (other approaches also exist).

But this has nothing to do with how you compute the integral. The properties of integration an differentiation are just different. That may indeed be due (as HallsofIvy said) to the more straightforward definition of the derivative, and the more complicated definition of the integral (whether Riemann integral or Lebesgue integral).
We can play semantic games, and "define" the derivative as the "anti-integral". But that wouldn't change the properties of integration, and wouldn't make computing indefinite integrals easier or more difficult.

The key are the properties, however you define the notions. The derivative of elementary functions can be computed algorithmically, and will yield an elementary function. Integration has nothing similar as property, so we are left with the fundamental theorem of calculus, and a bag of tricks (substitution, integration by parts,...) to compute an integral.

EDIT: not an answer to your question, but you may be interested in the Risch algorithm.

Last edited: Feb 29, 2016
10. Feb 29, 2016

### Staff: Mentor

This...

I agree that it wouldn't make any difference which operation you defined as the inverse of the other.