# Abuse of Liebniz Notation

1. Jun 7, 2009

### IniquiTrance

Why is it that performing algebraic operations on differentials in Liebniz notation is considered an abuse of the notation?

In every case, it seems like the operation makes sense to me.

Chain rule etc:

$$\frac{dy}{du}\frac{du}{dx} = \frac{dy}{dx}$$

Since differentials are just limits, if they exist, then they're regular fractions.

Thanks!

2. Jun 7, 2009

### slider142

No, they are not. Depending on the actual terms in the limit, the behavior of a limit may be quite different from the algebraic form of the term being considered. Consider the example in this post, which a student may mistakenly believe to be correct analysis inspired by the ease of Leibnitz notation. Do not trust the notation too much; always check the analysis.

3. Jul 3, 2009

### chrisr999

Derivatives ARE fractions.
By definition, a derivative is the slope or gradient of the tangent to the graph. The tangent is a straight line and the limit is necessary to find the gradient of the tangent rather than a line cutting through two points on the curve in the graph!!
How we combine derivatives is as follows...
suppose we want the derivative of {(2x+3) squared} with respect to x. We simply multiply the derivative of {2x+3 squared} with respect to 2x+3 by the derivative of 2x+3 with respect to x,
because it is the same as the derivative of {2x+3 squared} with respect to x.
We are "multiplying by 1".
It can be performed within "first principles notation" or the shorter notation of Leibniz and co.

Last edited: Jul 3, 2009
4. Jul 3, 2009

### g_edgar

When you try the same thing for changing variables in double integrals you will be in trouble. To change from dx dy to du dv you have to multiply by the Jacobian.

P.S. Leibniz not Liebniz

5. Jul 3, 2009

### chrisr999

of course, that is integration, the calculation of area, volume etc,
in the forum question, we simply multiply by 1 in the form of dy/dy, dq/dq and so on and swop around the denominators as in this situation (gradient slopes), we are undoubtedly dealing with fractions (ie 10/3 times 6/5 is 10/5 times 6/3 equals 4 as both are 60/15). That is the simple technique where that technique is applicable!!

Last edited: Jul 4, 2009
6. Jul 3, 2009

### chrisr999

Slider,
If x+y+z=0 then x=-(y+z), y=-(x+z) and z=-(x+y).

dy/dx is -1-dz/dx
dz/dx is -1-dy/dx
dx/dz is -1-dy/dz
dy/dz is -1-dx/dz
dx/dy is -1-dz/dy etc

and so on, since if any of x or y or z change, the others must change such that they still sum to zero and we must rule out the possibility of one of the three being a constant, which has not been specified by the equation, though you say they are all variables but the nature of their variation must be specified. You cannot state the derivatives the way you have done as you are suggesting x does not affect y, z does not affect x and y does not affect z which is contradictory

7. Jul 3, 2009

### jgens

You're clearly unaware of partial derivatives, especially since slider's analysis (Hurkyl's actually) is entirely correct. Derivatives are not quotients, they are limits of quotients. Without the limit concept, derivatives are useless fractions of the form 0/0. Because of this fact that, we cannot treat them as fractions in any rigorous sense.

8. Jul 4, 2009

### snipez90

Well if you are just taking derivatives, I don't really see a problem. I don't think anyone would compute a derivative incorrectly because he did not understand the notational peculiarities.

Now if you were to prove anything, an appeal to Leibniz notation is probably not a good idea. This is pretty clear I think. As for Leibniz notation itself, I think the abuse of notation is exaggerated. Again, computationally, I doubt many people would be lead astray by intuition, as long as they realize that it is pretty hard to be rigorous about infinitesimally small quantities.

But still, the OP probably made the most important observation regarding the continued use of Leibniz notation. Leibniz notation is popular because it is flexible and useful. The operations should make sense because Leibniz notation allows us to treat dy and dx in exactly the same way as if they were ordinary numbers. This is why it is so handy in calculations. But whatever, taking a derivative is kind of boring anyways.

Last edited: Jul 4, 2009
9. Jul 4, 2009

### chrisr999

the context assumed is "derivative as the tangent gradient".

0/0 is not a fraction! zero is not a number!
When calculating a derivative, we must find a way to "cancel" the infinitesimally small quantity in the denominator, typically for instance by locating it as a factor in the numerator.
As an example, what is {x(squared)-9}/{x-3} when x=3?
It is the fraction X+3 written in a deceptive way.
The trouble with 0/0 is this.....
How many zeroes is the numerator and how many zeroes is the denominator?
since 0(1)=0(2)=0(3)=0(4), keep doing nothing and end up with nothing.
It's undefined because we have lost the details of how many times the denominator is the numerator.
Even though derivatives are calculated with limits, since we are dealing with tangents to a single point, not a line through two points.... if the limit exists the derivative is a line gradient which is a quotient, no matter how we refer to it.
It works where it is applicable and not where it is not applicable. If the partial derivatives are across different unrelated variables it won't work!
We cannot divide miles by hours and give our answer in km/hour without converting correctly first. Slider's analysis is incomplete. he needs to state more clearly the relationship between x, y and z.
When performing partial differentiation, you are holding one of these constant. This is different to finding the derivative of a function of a function (say x).
Mathematics has many areas that seem the same but we may need to be more specific.
For instance CosA is either the ratio of two sides in a right angled triangle or the x co-ordinate of a point on the circumference of a unit radius circle, computed using the right-angled triangle.
In the context of the Chain Rule of differentiation, we are only "multiplying by 1" in a way that suits us, similarly to how we combine fractions with different denominators. If the word "derivative" means something else in a slightly different context, we are playing a different game of football.
When calculating the volume of a cube, the variables are "unrelated". Width, height and depth are "independent" unless restrictions are placed on the volume. Under these circumstances dx, dy and dz may not have any interrelationship as they are infinitesimal lengths independent of each other.
Partial derivatives are "partial", used in situations where the function depends on more than one "independent" variables. it's a different ball game. For the variables that are independent, there is no derivative as we cannot draw of a graph of one versus the other.
In Slider's analysis he obtains three partial derivatives multiplied together to get -1 when we'd expect 1 if we were dealing with derivatives. This is because he is calculating partial derivatives, not derivatives, by holding a variable constant.
A partial derivative is not a complete derivative. It is a slight redefinition of a derivative to deal with multivariable situations, a variation of a context.

Last edited: Jul 4, 2009
10. Jul 4, 2009

### HallsofIvy

And, let me point out, it is quite possible for a function of several variables to have partial derivatives at a point where it is NOT differentiable.

11. Jul 4, 2009

### chrisr999

It is true that we cannot treat derivatives as fractions in cases of multiple variables such that (dy/dx)(dx/dz)(dz/dy) is not equal to one, in other words, the "numerators" and "denominators" do not cancel as fractions normally do, but this is due to being in a three-dimensional space.
In the situation where we have a function of a function of x, the Chain Rule extracts the derivative of a composite function by multiplying by 1 and juggling the denominators.
In the case of multiple variables however, the partial derivatives are fractions but "not as we know them"!
In other words, they are not "combinable as fractions" until we clear up what kind of fractions they are, similar to the 0/0 case which has hidden the details of the fraction.
In the partial derivatives case, since we have not simply multiplied dy/dx by du/du to get (dy/du)(du/dx) by swopping the denominators, allowing us to differentiate this type of "composite function", we do not have the liberty of swopping around the numerators and denominators. Our context has changed.
To treat the notation as fractions, the context needs to be checked to see if the numerator symbol is really equal to the denominator symbol in another fraction!
I think the thread was started assuming functions of a single variable, however the introduction of the partial derivatives shows the importance of knowing the context.

Last edited: Jul 4, 2009
12. Jul 4, 2009

### Hurkyl

Staff Emeritus
No, they're not. (First derivatives in one-dimensional parameter spaces) are merely similar to fractions. e.g.

you have this backwards. We don't swap denominators because derivatives are fractions -- we say derivatives are similar to fractions because we've proven by other means that you can "swap" denominators.

One of the dangers of lying to students and telling them derivatives are fractions is that they might believe you -- and they'll start doing all those other things you can do with fractions that don't work here.

e.g. they'll immediately start to wonder what dx all by itself would mean. If they figure out the idea of a differential form... wonderful! But if they don't and get some other idea stuck in their head because we pointed them down that road and gave them no guidance... bad.

Or another example: if derivatives are fractions, then obviously the following should be an identity, right?
$$\frac{d^2 f}{(dy)^2} \cdot \left( \frac{dy}{dx} \right)^2 = \frac{d^2 f}{(dx)^2}$$

There are wonderful concepts available in Calc I for the students to learn -- e.g. things like the idea of a first-order approximation. Pretending we're just doing fancy things with fractions would (appear to) obstruct their chances of learning that.

Last edited: Jul 4, 2009
13. Jul 4, 2009

### chrisr999

it would pay to clear up the language.
A derivative is a fraction because it mathematically calculates the gradient of a tangent to a curve, that's a fraction since a line gradient is (delta y)/(delta x).
however, if you are operating with different variables, the fractions may not be compatible!
that's clear,
derivatives work the same way as fractions when you have compatibility. don't forget dy/dx and so on is a notational short cut and is really tan(angle) for the tangent.

derivatives are fractions because a line gradient is.
BUT there are situations when you can use the fraction math because we are dealing with compatible fractions. there are obvious cases where you can't.
If students learn what is actually being calculated, they can learn without ambiguity.
Telling them derivatives are not fractions is only adding to the confusion.
There is far more to the picture.

Last edited: Jul 4, 2009
14. Jul 4, 2009

### chrisr999

That is being way too obvious, Hurkyl,
of course that situation is not a pair of fractions where you can simply cancel notational terms. A dialogue should clear up confusion.
One of the difficulties with calculus is the notation.
It was developed from sound principles using straight lines, whereby a gradient has a numerator and denominator, leading to a value evaluated at a limit as the denominator approaches zero.
In certain situations, when dealing with a single variable, the "quotient" can be dealt with as a fraction if you work it through in first principles form. however you cannot say in general that we can simply cancel "like symbols" such as dy, dx, dz etc.
It all depends on the situation being analysed.
You cannot exclusively say that derivatives are not fractions, nor can you exclusively say that a dx in a numerator position must cancel with a dx in a denominator position.
They are much more subtle fractions.
It would pay a student to understand the different cases.

Consider the graph of y=x(squared), or similar (single variable).
if you draw the graph and pick a point x,y on the graph, then if you draw the tangent, you can view dy/dx or d(f[x])/dx using delta(y)/delta(x) for the tangent itself.
if you turn the page of the graph to it's side, you can now view dx/dy or dx/d(f[x]), which is the gradient to x = f(y) at the same point (x,y). One of these is a/b and the other is b/a.
In situations of a single variable like this (dy/dx)(dx/dy) is 1 as we have (a/b)(b/a).

It is not that dx is the numerator and dy the denominator!!!
That is notation!! written like a fraction for a reason.
The ratio of dy to dx however is tan(angle) for the tangent.

Last edited: Jul 4, 2009
15. Jul 4, 2009

### Hurkyl

Staff Emeritus
You say "of course", but it's not going to be obvious to the student -- especially if you're telling him derivatives are fractions. (and students aren't the only people who make this mistake!)

Yes, that's right. And the dialongue should emphasize to the student that derivatives aren't fractions, despite having some similar arithmetic properties.

If you want to redefine the word "fraction" to include both fractions and derivatives, that's fine -- but you need to make sure the student understands that you've done that. (aside: I really don't see why using the same word for both concepts should be preferable)

16. Jul 4, 2009

### chrisr999

Students learn by reference to previous learning.
Then the student can say "this is like that but different in this new way",
by referring to previous knowledge. Otherwise, Mathematics becomes disjointed, learning becomes disjointed and when that happens, we have the ground for confusion.
Mathematics is a universal language, though we may have various ways of wording it.
Understanding the similarities and dissimilarities generates comprehension, breakthroughs in understanding.
I've seen a student totally bewildered by the statement in a textbook that "derivatives are not fractions" and began to struggle with the Chain Rule.
They may not be constant/constant due to the changing direction of the graph but derivatives are tangent gradients.
If we start saying derivatives are not fractions, then we should clear up what we mean, we are certainly not saying a derivative is anything other than the gradient of a tangent, and the gradient of a tangent is delta(y)/delta(x). We cannot choose two points on the curve to find this but we can choose any two points on the tangent.
Students first learn with averages in school. The speeds they work with are average speeds, then we introduce calculus to show how we calculate instantaneous speed by letting the time interval approach zero, graphically speaking we are bending a line until it is a tangent to the graph of distance versus time. By finding the limit (as delta t goes to zero) we are discovering the gradient of the tangent, which of itself is a quotient representing distance/time at an instant in time.

17. Jul 4, 2009

### chrisr999

Although we can call dy/dx "notation",
it's really tan(angle) of the tangent to the curve. It may be impossible to see at the limit, but if you magnify it greatly, then it is the right-angled triangle flush against the tangent, with the perpendicular sides parallel to both axes.
That's what we mean by calling it a fraction.
dy/dx is delta(y)/delta(x) for the tangent, or opposite/adjacent, since proportions are maintained at any scale for equiangular triangles.

dy/dx for the curve is delta(y)/delta(x) for the tangent. For a single variable, y=f(x) then for a point (x,y) derivatives on either axis can be combined as fractions.

18. Jul 4, 2009

### Hurkyl

Staff Emeritus
Students haven't learned about differential forms or number systems with nonzero infinitessimals yet or whatever technical idea is needed to make sense of your exposition. Your attempts to "explain" the derivative is not "referencing previous learning". Instead, you are introducing new ideas and coercing the student to think in terms of the new ideas, despite the fact you are going to give them absolutely zero guidance in the understanding of those ideas.

Yes, there is previous learning to reference. One has the idea of tangent lines -- and (AFAIK) the first thing most textbooks do to introduce derivatives is the heuristic method of computing the tangent line as the limiting secant line.

(I put limit in bold because it's really important, but it's something you appear to be downplaying)

One also knows of inequations, in terms of which the important idea of differential approximation can be introduced. (I assert that differential approximation is by far the most important concept in Calc I)

There's also the idea of "rate of change", but the more I think about it, the more I think that really ought to be delayed until after the fundamental theorem of calculus, so that the student can be shown how the "rate of change" relates to actual changes.

19. Jul 5, 2009

### OrderOfThings

This is a correct identity. The fractions $d^2 f/dx^2$ and $d^2 f/dy^2$ equal the second derivative of f only under the assumption that $d^2x$ and $d^2y$ vanish. The general formula for the second derivative of f with respect to x is
$$\frac{d^2\!f\,dx-df\,d^2x}{dx^3}$$

20. Jul 5, 2009

### chrisr999

I might post a diagram to explain what I was trying to explain Hurkyl,
the person who started the post wanted to understand how derivatives relate to ratios.
That can easily be shown using a tangent at a specific point and showing that dy/dx is a regular ratio at that point and the value of the ratio varies as you move around the graph.
At a specific point derivatives can be worked as fractions.
a clear geometric analysis can show this.
If you want to help that's ok but derivatives are pretty much geometrical creatures and the notation has to be developed around that