You CAN multiply by dx! #SOMEpi

dy/dx is a ratio and you CAN multiply and divide by dy, dx and so on like a FRACTION!!! You don't belive me? Well, I can assure you that they are numbers and it's completely rigorous to treat them as a fraction (ratio). In this video we'll see how it can be done and the other ways in which that dy/dx symbol can be interpreted (at the end all identify the same object even though they aren't the same thing!). Like d/dx as an operator (or functional) that acts on some function or as a simple notation that just means the derivative with respect to the variable x!!!

#SOMEpi #calculus #math
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[JonnyMath]

Did you like it!!!🤩🤩🤩 I've finally (or at least I hope😅🤣) ended this debate!!!😅🤣😉🤗

[mathfrommydesk]

This property is the reason why the Leibniz notation is my favorite

[MoiraLachesis]

Nicely accessible explanation of differential forms

[ruzgar]

f’(x)h = f(x+h)-f(x)

h = dx
f(x+h)-f(x) = dy

[magma]

This is actually how differential forms works, where we define the derivative operator as dx_{n} where each x_{n} is a basis vector of sorts and dx_{n} is also a basis vector of sorts, but over an exterior algebra, and the wedge product is takes two elements and produces an orientated area. If we only have one basis vector, we would end up with df=f’(x)\wedge dx, which is effectively saying that dx^{2}=0, meaning that we can replace the wedge product with standard multiplication, giving us df=f’(x)dx, we can also integrate over a surface, giving us \int_{Ω}df=\int_{Ω}f’(x)dx which is basically the anti-derivative.

[samueldeandrade]

Can someone tell me who started to interpret "fractions" and "ratios" as different things? It makes no sense to do that!!! And we actually do NOT do that already. There is a thing called

Field of Fractions of an integral domain

So, please, tell me, who was the genius that decided it was a good idea to use "ratio" for

π/e, (x²+1)/x, etc ... ?

Also,

2/3 is a fraction, but 2k/3k is a ratio, for k non-integer ...

Wow, that's good terminology.

[ProjectionProjects.]

I really don't get why people try to argue that this interpretation is some how "Invalid" or "non-rigorous". I get that the epsilon delta definition of the derivative exist, but who said we only can have one way of thinking of the derivative?

[cleon_teunissen]

About the claim: this is rigorous.
Let me make a comparison with another context: if you have a differential equation that on one side specifies taking the square root of that whole side, then the first step to begin solving that equation is to square both sides. That operation is likely to introduce a spurious solution. Chances are you end up with two solutions, and you have to check against the original equation to see which solution satisfies the original equation.

Returning to dy/dx: The concern, I assume, is: shifting between treating as a ratio and treating as an operator may under specific circumstances allow a wrong result to slip in.
The example that you discuss shows that the method will give the correct result in many cases. The question is: under any circumstances, is it certain a wrong result cannot slip in?

As an example of a wrong result that slips in: "proofs" that 1=0

[APaleDot]

I'm not sure what the point of the middle of the video was. Showing that dy = dx only works if y = x. And then you didn't use that equality anywhere anyway. Seems to just confuse the matter.

[pietrocestola]

In the one-dimensional case, when you form the quotient

df(x, Δx)/dx(x, Δx) = f'(x)Δx/Δx = f'(x),

the Δx cancels, leaving an expression that depends solely on x. This cancellation guarantees that the derivative is independent of the particular increment chosen (as long as Δx ≠ 0).

In the multivariate case, however, consider the expression

df(z, Δz)/dx(z, Δz) = ∂f(x, y)/∂x + (∂f(x, y)/∂y)(Δy/Δx).

Here, the quotient still contains the ratio Δy/Δx, which means that the result depends on how Δy relates to Δx. This directional dependence prevents the definition of a unique derivative using the quotient alone. Instead, the derivative in ℝ² at a point must be defined as a linear map.

[MoiraLachesis]

Ok there are multiple errors in the solution of y' = y but they're nbd for this context.

1) you have to consider y = 0 before dividing (it's actually a solution)
2) the integral of 1/y can have two different integration constants for positive and negative y
3) |y| > 0 does NOT imply y > 0
4) If you pull out c from the exponent the factor C = e^c would be positive.


Amusingly, dropping zero and negative solutions then forgetting that C is positive reintroduces the lost solutions and yields the correct result.

[Xponent-nbhe]

wait you’re left handed? me too, I considered getting myself a whiteboard for fun but I thought I would definitely smudge all over the board when I write lol (nice video btw)

[RossMiller-gd]

Your explanation is confusing, how can you have dy with delta x present when dy is dependent on delta x approaching zero ?

[coreymonsta]

So the reasoning is the fact that you found someone else call this dx a number and then you did some calculations lol

[kirbo]

I've never went further besides this, but now it's clear for me that it is in fact rigorous to treat dy/dx as a ratio🎉👍

Also, what program did you use for the thumbnail?👀

[romaing.]

I thought this was going to be a video about differential forms... You should take the habit of researching a bit more when you wonder about something, as often many people had the same idea before and developped it.

[lukfiorentino]

I do not agree with the statement: we can assume the differential of the identity function equals the horizontal increment taken for measuring the linear increment of the function, there are orders of infinitesimals and infinites therefore I could accept that for x that tends to 0 sin(x) is asymptoticaly equivalent to the identity but I would not say the same for x that tends to infinity

[raelerminy]

Everything ok, but an entire explination requieres talking about differential forms.

[HadiLq]

It's not regoreus! dx on its own is zero. If you forget it's all about limits, where you will in this way, it's easy to make nonsense out of it.

[djbar]

dx, dy --> infinitesimals - no dimentional value.
Delta x, Delta y --> finite diferences (numerical values).
Don't mix them, dy/dy is a limit, dy and dx are useful abstraction tools.
Perhaps you have no knowledge of numerical analysis yet.