# Differential by another differential

• Nick R
In summary: I don't think that there would be exactly one real number in the halo of a limited hyperreal)In summary, the next step in the derivation of the product rule is to consider dudv = 0 because, when taking the ratio with respect to an infinitesimal change in x, dudv becomes negligible and approaches 0. This is similar to how a ratio of a real number and a non-zero real number approaches 0.
Nick R
In deriving the product rule,

d(uv) = (u + du)(v + dv) - uv

= udv + vdu + dudv

The next step in the derivation is to consider dudv = 0 because it is "negligible". I have seen this elsewhere too.

I can see that dudv would be a very small number, but why is it valid to say that dudv = 0?

See if this helps: http://web.mit.edu/wwmath/calculus/differentiation/products.html"

Last edited by a moderator:
Nick R said:
In deriving the product rule,

d(uv) = (u + du)(v + dv) - uv

= udv + vdu + dudv

The next step in the derivation is to consider dudv = 0 because it is "negligible". I have seen this elsewhere too.

I can see that dudv would be a very small number, but why is it valid to say that dudv = 0?

It is only zero when the limit is taken.
the differential is defined as
dy=lim Δy
it is know from finite calculus
Δ(uv)=(u+Δu)(v+Δv)-uv
=uΔv+vΔu+ΔuΔv
lim ΔuΔv=0
but ΔuΔv is not 0
clearly if Δu and Δv are each o(Δx)
ΔuΔv is o((Δx)^2)

Nick R said:
In deriving the product rule,

d(uv) = (u + du)(v + dv) - uv

= udv + vdu + dudv

The next step in the derivation is to consider dudv = 0 because it is "negligible". I have seen this elsewhere too.

I can see that dudv would be a very small number, but why is it valid to say that dudv = 0?

That proof, while not a bad "rough" way of seeing what is happening, is only valid in "non-standard calculus" where you have first given a rigorous definition of "differentials" (which itself requires some pretty deep logic in order to extend the real number system). IF you had done that then you know that, from the definition of differentials, the product of any differentials is exactly 0.
Better to stay with the "more complicated on the surface but simpler in essence" proof using limits.

HallsofIvy said:
That proof, while not a bad "rough" way of seeing what is happening, is only valid in "non-standard calculus" where you have first given a rigorous definition of "differentials" (which itself requires some pretty deep logic in order to extend the real number system). IF you had done that then you know that, from the definition of differentials, the product of any differentials is exactly 0.
You're confusing nonstandard analysis with differential forms (and/or dual numbers).

The differential of a function f is a function of two variables defined by df(x,h)=f'(x)h. The product rule for differentials follows from that definition and from the product rule for derivatives. No need to let h be small, or to take any limits.

What (branch or topic in mathematics) defines the operator d() and the associated "differentials" such as dx?

Depends on exactly what you mean. I interpreted what you said a treating the differentials directly- i.e. in "non-standard analysis" and was taken to task for it. If you are referring to the differential defined as Fredrik says (I would write the notation a little differently) it should be covered in any Calculus text.

HallsofIvy said:
Depends on exactly what you mean. I interpreted what you said a treating the differentials directly- i.e. in "non-standard analysis" and was taken to task for it. If you are referring to the differential defined as Fredrik says (I would write the notation a little differently) it should be covered in any Calculus text.

I was under the interpretation that if du and dx are infinitesimals (with respect to the real numbers), then the galaxy of dudx does not contain either du or dx (i.e., dudx is infinitesimal with respect to both du and dx), but because the Hyperreals are a field, if du and dx are both non-zero, then dudx is non-zero.

Sorry, I should define some of those terms:

A Hyperreal y is infinitesimal with respect to x if there for all real numbers r
|ry| < |x|

Similarly, x is said to be unlimited with respect to y if the above holds.

A number is said to be infinitesimal if it is infinitesimal with respect to the real numbers, and a number is said to be unlimited (or infinite) if it is unlimited with respect to the real numbers.

The halo of a number x is the set of all numbers y such that x - y is infinitesimal with respect to x (I might be wrong about this... it might be that they are infinitesimal with respect to the real numbers)

The galaxy of a number x is the set of all numbers y such that x - y is not unlimited with respect to x (again, I might be wrong about this, it might be that they are not unlimited with respect to the real numbers, but in my post I was using the definition that they are not unlimited with respect to x)

Since X=aR where a is some real number and R is any real number (x is some multiple of any real number), I'd think definitions for galaxy and halo must be with respect to any real number.

Looking at those definitions, maybe this is why dudv is considered 0 in the derivation...

say you are differentiating uv with respect to x

d(uv) = (u + du)(v + dv) - uv

= udv + vdu + dudv

but we want the ratio of this with respect to an infintesmial change in x so we divide by dx

dividing...

d(uv)/dx = u(dv/dx) + v(du/dx) + (dudv)/dx

and since dudv is infinitesimal with respect to dx, or, |r(dudv)| < |dx| ; r is a real number

(dudv)/dx = 0

similar to how

Lim(a -> 0) [ a / b ] = 0 ; b is some real number != 0

Last edited:
So for any limited hyperreal number (anything that is not infinite with respect to the real numbers), there is exactly 1 real number in its halo

(Also, now that I think about it, i do think that the definition of galaxy and halo are with respect to the real numbers because otherwise, the only number in the halo of 0 would be 0, but with the usual definition, the halo of 0 is every infinitesimal number. In that case, what I meant is that dudv is infinitesimal with respect to du and dv, so you can ignore the stuff about galaxy and halo there)

The only number real number in the halo of an infinitesimal number is 0, so if we take the real number closest to d(uv)/dx, you get uv' + vu'
where v' is the x derivative of v and u' is the x derivative of u
(since the real number closest to (dudv)/dx is 0, that part disappears)

Last edited:

## What is the definition of "differential by another differential"?

The differential of a function is the ratio of the change in the output of the function to the change in the input. When we take the differential of a differential, we are essentially finding the rate of change of the rate of change.

## How is "differential by another differential" related to derivatives?

Differentials and derivatives are closely related concepts. The differential of a function is, in fact, the derivative of that function multiplied by the change in the input. So when we take the differential of a differential, we are essentially taking the second derivative of the original function.

## What is the notation for "differential by another differential"?

The notation for taking the differential of a differential is d²/dx² or d²y/dx², depending on the notation used for derivatives. The "d²" represents taking the second derivative, and the "dx²" or "dy/dx²" represents the change in the input.

## What is the purpose of taking "differential by another differential"?

Taking the differential of a differential is useful for understanding the rate of change of a function. It can help us determine the concavity of a function, or the shape of the curve. It is also used in various equations and applications in physics and engineering.

## Can "differential by another differential" be applied to any type of function?

Yes, the concept of taking the differential of a differential can be applied to any type of function, whether it is linear, quadratic, exponential, or any other type. However, the notation and techniques used may vary depending on the type of function.

• Calculus
Replies
2
Views
912
• Calculus
Replies
22
Views
2K
• Calculus and Beyond Homework Help
Replies
21
Views
1K
• Calculus
Replies
2
Views
1K
• Calculus
Replies
4
Views
2K
• Calculus
Replies
4
Views
978
• Calculus
Replies
3
Views
2K
• Calculus
Replies
6
Views
2K
• Calculus
Replies
2
Views
1K
• Calculus
Replies
4
Views
3K