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@Frankiethedogishere

11 күн бұрын

As a dog, I don’t understand! Howevere, I am first pinned comment replier! 🐕😆

"Nooo don't treat dy/dx as a fraction it only works 100% of the time"

@user-iv4dh7zp7s

21 күн бұрын

well at least dy/dx isn't always y/x by cancelling out the d's

@cubicinfinity2

21 күн бұрын

@@user-iv4dh7zp7s lol yes.

@davidbrown8763

21 күн бұрын

My thinking exactly. I believe that it works because it IS a fraction. I see it as the rate at which y is increasing with respect to the increasing rate of x at the infinitesimal level. (Please note that "rate of increase", can also be negative - in which case the variables are decreasing at the infinitesimal level). But what do I know?

@flewawayandaway4763

20 күн бұрын

It doesn't work in case of double derivatives

@vignotum132

19 күн бұрын

I’ve heard there are cases where it doesn’t work

The reason it works is because it IS intended to be a fraction. Indeed, Leibniz thought of dy/dx as a ratio of infinitesimals. So it's by no means some coincidence that this works, like some people seem to think. Actually, theories of infinitesimals that support calculus do exist, as shown by Abraham Robinson in the 1960s with his hyperreal numbers. If you are interested, there is a good entry-level book by H. Jerome Keisler called "Elementary Calculus: an infinitesimal approach".

@kwan3217

26 күн бұрын

I was here to say the exact same thing, even with the reference to the same book. I'll take every opportunity to spread the gospel of infinitesimals.

@myboatforacar

26 күн бұрын

Please... that book is barely anything ;)

@martimzurita

25 күн бұрын

What a beautiful book this one by Keisler, thanks so much for sharing it!

@_ranko

23 күн бұрын

Lmao infinitesimal approach is non-standard for a reason, they encountered a bunch of problems with defining calculus with infinitesimals so they dropped it in favor of limits. Doesn't matter what it's intended for, it's objectively wrong in the de facto definition

@naytte9286

23 күн бұрын

​@@_ranko the reason non-standard analysis is non-standard is because the logic required to develop calculus rigorously with infinitesimals did not exist up until recently. Łoś's Theorem is arguably what allowed Robinsson to develop his hyperreals, and that theorem was not formulated until the 1900s, that is, after real analysis was already developed. Also, I didn't claim that dy/dx can be literally interpreted as a ratio in standard frameworks, simply that dy/dx was INTENDED, by Leibniz, to be a ratio.

In physics it's common practice to treat Leibniz's as a fraction.

@kacpermalinowski8215

26 күн бұрын

Physicists will also do this, so idk if that's a good argument kzread.info/dash/bejne/gK5_k5N9orXKm9Y.htmlsi=uexzXJEVOzibOdco

@ohlala2219

26 күн бұрын

Yeah they always do these sort of things to math xD

@intellix7133

26 күн бұрын

It feels like a cheatcode honestly, but I'll gladly accept it to compute my damn integrals

@juanmanuelmunozhernandez7032

25 күн бұрын

It cannot be a coincidence every single time that it works to give experimentally demonstrable theoretical results. Intuitively, it's got to do with some regularity properties when taking limits of increments into derivatives and integrals

@alessandrocoopman9135

25 күн бұрын

@@juanmanuelmunozhernandez7032 Velocity is defined as "the rate of change of position with respect to nonzero time" and its unit is meters per second (m/s). Then it's natural to treat it like a ratio. Therefore, all derivatives that follow in physics are fractions (better, ratios), even if you cannot strictly "divide" numerator by denominator.....weird isn't it?

Whenever the d is STRAIGHT and FIRM, you can cancel. Whenever the d is flaccid - I mean curved, you know, partial derivatives - then you can’t cancel.

@mosescheung5794

26 күн бұрын

the what is WHAT

@themathematicstutor4092

26 күн бұрын

Taking the d with respect to x

@nestorv7627

26 күн бұрын

I love phallic math

@frederf3227

20 күн бұрын

Unless you d isn't simple drivative but total derivative in a multivariable case

@whannabi

17 күн бұрын

​@@nestorv7627When the d has ED, you stay away because it's flaccid

I am an Engineer, I see dy/dx and chances are treating it as a fraction is part of my solution.

@johnjameson6751

22 күн бұрын

And it will always work, because dy/dx is a ratio of differential 1-forms on the real line (or an interval).

@v12-s65

9 күн бұрын

@@johnjameson6751for small approximations mostly 😜

@SoloRenegade

8 күн бұрын

change in y, over change in x It's literally a fraction It's literally the slope of a line

@ingGS

7 күн бұрын

@@SoloRenegade In the sense that you and I use it perhaps, but tell that to 1600-1700 mathematicians and debate was sure to come your way. Calculus was just thought to be not rigorous enough, the concept of infinitesimals was not widely accepted, and more importantly what the heck would the ratio of this "concept" be? It took all the way to Cauchy (19th century) for it to be properly formalized.

@SoloRenegade

7 күн бұрын

@@ingGS doesn't matter. literally nothing you said disproves it being a fraction. even today, infinity doesn't exist. never once has an infinity of any kind ever been observed in the universe, not even in math.

So then, why does the chain rule work? Turns out, the proof of the chain rule is verbatim (with a well-definedness check so that we don't divide by zero) going to the difference quotient and performing a cancelation of fractons before taking a limit. Yes, dy/dx is a fraction. It's just an infinitesimal fraction; that requires developing careful intuitions for it. Ones which btw do not hold in multiple variable contexts: partials behave quite differently sometimes from their one-dimensional cousins.

@angeldude101

26 күн бұрын

I've yet to see a case where partial derivatives are meaningfully distinguished from ordinary derivatives where the derivative of one input with respect to the others is assumed to be 0.

@feuerwelle4562

26 күн бұрын

dy/dx is not a fraction. It can "behave" like a fraction and this way of thinking might be good for intuition, but it's really not a fraction.

@lauren_e_s

26 күн бұрын

It's not a fraction, it's a limit of a fraction

@alb2451

26 күн бұрын

@@feuerwelle4562 d(fog)/dx = df/dg * dg/dx = lim(u->0) f(g(x+u)) - f(g(x))/g(x+u) - g(x) * g(x+u) - g(x) / u. The terms cancel out, they're fractions.

@ElectroNeutrino

26 күн бұрын

@@alb2451 That's a limit of a fraction, not necessarily a fraction, since the divisor is going to zero. Treating it like a fraction is abuse of notation. Edit: That being said, it's still a useful approach that gives the same results as a more rigorous treatment would. You'd need to turn to non-standard analysis in order to rigorously treat it like a fraction.

Thats gotta be the most anti-algorithm title ever

@SeeTv.

23 күн бұрын

What do you mean? It works really well in combination with the thumbnail("This isn't a fraction: dy/dx"). The thumbnail is part 1, the title is part 2.

@bhaveerathod2373

23 күн бұрын

I was just thinking that 😂

@uKaigo

22 күн бұрын

​@@SeeTv. It makes sense to us, not for the algorithm. The algo may use OCR (I don't know if it does) or voice recognition/subtitles to identify the topic, but the title alone + description does not. Hence, it's anti algorithm.

@Marstio1

14 күн бұрын

Yet it still appeared on my yt page. Though I’m a comsci major, but I don’t really watch these types of videos much

@uKaigo

14 күн бұрын

@@Marstio1 I mean, the algorithm does not take only 3 things into account. Though the algorithm is a black box so we can't know fore sure It was also recommended to me This could be because multiple people with similar interests watched this video, so it assumed we would too But, it's hard to search for the vídeo (I couldn't but that might be a skill issue)

Hi Aerospace Engineering Lecturer here. We are a special breed in Engineering where, we use the egghead's mathematics but are 100% utterly honest and forgivingly blunt about it. When the power hour duo of the Issacs (Barrow and Newton) both both began developing the more modern derivative by limits definition, it has been hailed as this beautiful thing in math with the complicated name that gives students headaches. In engineering we teach it way more bluntly. The Issacs used the most beautiful tool in human beings tool kit. Their inner morons 😆Like actually think about it... "I cant measure the slope of a curvey line! SOD IT WE DOIN IT LIVE BOIS!" *Proceeds to slap a straight line on the curve and force it into submission* My personal theory as to why, over the years, we've said that dy/dx is not a fraction is simple... Its the same reason we decided to call the theory in beam bending that allows you to just add 2 beam cases together (literally the theory, "Addition go brt") the "Principle of Superposition". We don't like admitting our "Smart" ideas are actually much dumber than we pretend. We are the same species that came up with 3 laws of thermodynamics... then found a 4th but it was the most important and went "Should we change the 1st to 2nd, 2nd to 3rd, 3rd to 4th and make this new one 1st? Sod it call it the 0th law... dy/dx is not a notation, I adamantly will fight any mathematician on this... dy/dx is the brutally honest admission of what we have done to get our fancy differential equations... We slapped a straight line onto a curve and measured its gradient after making the line infinitely small... And the equation for a gradient of a straight line? y2 - y1 / x2 - x1 aka.... ChangeY/ChangeX aka dy/dx.... We just really REALLY don't like thinking about the fact that the entirety of one the most revolutionary and culturally seen as "most difficult and academically impressive" math we've done as species boils down to "hehe line on curve but small hehehehe". In engineering we arn't afraid to admit that we, beyond all our fancy text books and our latin-based names... are still just stupid monkey brains who have been bodge jobbing our way to success for centuries 😅 dy/dx is a fraction, we should start accepting that...

@kylenetherwood8734

8 күн бұрын

0th law was discovered last because it feels so obvious. You basically just need to say that temperature exists. Hence why it's stuck in front because it's so fundamental to the topic. We also did the zero thing with the time dimension. It's just a nice number.

@tikoblocks3224

8 күн бұрын

I like how you write

@DrWhom

7 күн бұрын

spoken like a true engineer 100% confused bullcrap

@aspartamexylitol

7 күн бұрын

​@@DrWhomcope

@solidpython4964

6 күн бұрын

Infinitesimals aren’t really rigorous the way Cauchy’s definitions for limits are, you should look at analysis more before asserting things so confidently. See some of Berkeley’s refutations of how Newton and Liebniz would use infinitesimals to do nonsensical things, until we had a more concrete understanding of a limit later down the line.

are there any examples showing that treating dy/dx as a fraction doesn’t always work? edit to clarify: with ordinary derivatives, not partial

@Arthur-so2cd

26 күн бұрын

yes, the simplest occured with me in physics class. Suppose F=F(x,y,z), then dF/dq = (partial F/ partial x)*(partial x/ partial q) + the same, but change x for y + the same, change x for z. If you cancel out all the partial x's and y's and z's you end up with 1=3

@_jktujq_7666

26 күн бұрын

Problem lies only in partial derivatives and someone has already mentioned it, but when there are upright d's, it is perfectly fine to treat those as fractions, for example take function y=f(t), dy/dx * dx/dt is exactly dy/dt (simple chain rule, which is mentioned in video).

@rarebeeph1783

26 күн бұрын

@@_jktujq_7666 yeah, and you should even be able to recover the nice behavior with partials by asserting that "partial F" on its own does not have a unique meaning. if we instead refer to each individual partial numerator as "d_x F", "d_y F", and "d_z F", then the fraction-like behavior seems restored to me. in particular, we may see the gradient vector [d_x F / dx ; d_y F / dy ; d_z F / dz] as equivalent to the pure differential dF = d_x F + d_y F + d_z F (that is, analyzing the basis vectors as being the denominators dx, dy, and dz), then we can divide by dq and multiply each term by 1, and recover the chain rule dF/dq = (d_x F / dx)(dx / dq) + (d_y F / dy)(dy/dq) + (d_z F / dz)(dz/dq) but i don't have any proofs about this so idk if it can be made rigorous

@munkhjinbuyandelger

26 күн бұрын

it doesn’t work in multivariable calculus

@lukandrate9866

26 күн бұрын

dy/dx is a fraction if you don't go higher than calc 2

I'm still not satisfied. The only complication seems to be because dy and dx are infinitesimal they have no useful value outside of their relation to each other. But treating dy/dx as a fraction and performing regular valid manipulations never breaks that relationship. As for people saying "well you can't cancel the d's", including my first year maths lecturer, that is completely stupid, of course not, the d's are not separate values or variables. (Which is why it is preferable to keep your d's roman or unitalicised btw.)

Looked in my calculus lectures and also in Wikipedia: The are different approaches to define differential operator. Our university stated for us that there are derivatives (y'(x)). Then that there's differential (d(y) = dy = y'(x)*dx). And there arises identity: dy/dx = y'(x). Wilipedia calls it Cauchy's approach (en.m.wikipedia.org/wiki/Differential_of_a_function).

But it's essentially the slope, which is a fraction.

We treat it as a fraction b/c it IS in fact a fraction: it's the ratio of two functionals, called differentials - these are sections of the cotangent bundle of the reals (the fibers just happen to be canonically the real line itself). If you think about it a little bit, the way to capture the original intuition of infinitesimal quantities in a rigorous way is precisely by using vectors (and consequently Linear Algebra) b/c on the one hand vectors are points, but on the other they are also quantities with magnitude and direction. Intuitively, if you let the magnitude get infinitesimally small, you are still left with the direction, so an abstract direction corresponds to an abstract (directed) infinitesimality (directed b/c for example it can have a sign).

@DrWhom

7 күн бұрын

exactly - basic stuff, right?

@maringenov7753

7 күн бұрын

@@DrWhom Right :)

"dy/dx is a fraction and most probably will always be a fraction to me. You can't change my mind."

I think of it as an infinitesimal change divided by another infinitesimal change.

@SergioLopez-yu4cu

24 күн бұрын

That's engineer mind.

@johnjameson6751

22 күн бұрын

The geometer mind is that it is the rate of change of y divided by the rate of change of x when moving with a given velocity. Thus dy/dx is a ratio of differential 1-forms.

@whannabi

17 күн бұрын

​@@johnjameson6751Geometry is underrated. It actually seems tangible/intuitive and no one will ask "how am I gonna use that in real life" because you can immediately see the applications. It's also a nice way to introduce to proofs because it mixes visual with notation but that's only a small list of what makes it great.

@DrWhom

7 күн бұрын

yeah, no

@usipussi6647

7 күн бұрын

Thats a fraction

According to Leibniz, the quotient of an infinitesimal increment of y by an infinitesimal increment of x. So it is a fraction

Bro really just proved the entirety of physics, my my

oh, so you include actual Brilliant content in your subject matter, meaning i can't just skip the sponsored part of the video? that's .... that's brilliant, damn you

I mean from first principles it really is just an infintesimal fraction, This is where all foundational derivative rules... including chain rule, come from.

@SergioLopez-yu4cu

24 күн бұрын

No, you can clearly prove chain rule doing valid operations on a limit, where have you coursed calculus, man? Leibniz defined this as a fraction and he was wrong, as simple as that. In his times he was discovering something new and of course there is a lack of rigor.

@hectormartinpenapollastri8431

24 күн бұрын

​@SergioLopez-yu4cu you can define derivatives as fractions rigurosly. You have to add mkre numbers to the reals that are treated as differentials, these are hyperreal numbers. The formulation of calculus along these lines is non standard analysis, and rewrites all calculus in a way that you can operate with differentials and the derivative is a quotient of differentials (literally).

@SergioLopez-yu4cu

24 күн бұрын

@@hectormartinpenapollastri8431 , in real numbers you can't, and most videos about this topic are obviously about real analysis and calculus, not non-standard analysis. If you are doing an exam of real analysis (or calculus), do you think you can just use numbers that don't belong to R and break the construction of that set? Of course, you can't, the definition of derivative in R is not the same as in *R, the same with natural numbers not being a subset of Z, we just assume that for convenience, but it isn't formally true.

@JoQeZzZ

23 күн бұрын

​@@SergioLopez-yu4cuof course you can use numbers that don't belong to R, as long as you use them as an intermediate value. This is the very essence of harmonic motion for example. Same goes for the hyperreals, in fact the definition of the derivative uses the standard part function, explicitly a *R -> R function. This means that whatever operations you do, you will always do R -> (...) -> R, and will thus constrain yourself to the real domain. If I derive the double angle formula using complex form of sin(x) and cos(x), is it suddenly not a R -> R identity anymore?

@SergioLopez-yu4cu

22 күн бұрын

@@JoQeZzZ , the definition of a derivative doesn't need *R, it's a limit.

This notation gets intoition from the world of differential forms where 0-forms are functions and taking exterior derivative of a 0-form f yields 1-form df=f'dx, where dx denotes the exterior derivative of the x-coordinate function (by coordinate function I mean we can generalize to R^n). So the notation comes from the equality presented above (in R^n sum of all partial derivatives df=\sum_i^n \partial_i fdx_i and is obviously generalized to manifolds).

@johnjameson6751

22 күн бұрын

Exactly: intuitively dy at each point is a function of velocity v, where dy(v) is the rate of change of y when moving through that point with velocity v. On the real line (or any interval), velocities are 1-dimensional, so two functions of velocity have a ratio, which is a real number and dy/dx is an example of such a ratio.

isn't it just a ratio? I mean we have differentiable function's rise formula: Δy = AΔx + o(Δx) where differential is defined by AΔx and it's just a product of A and Δx after doing some tricks with dividing by Δx and taking a limit we're getting f'(x_0) = A, so f'(x_0)Δx = dy and by this formula we can see that Δx = dx(not Δx ≡ dx) and thus f'(x) = dy/dx? I might be missing something, at least I think so. What did I get wrong?

@prant55

26 күн бұрын

so you're saying it's a fraction

@Endermanv-ot2if

26 күн бұрын

@@prant55 nooo you dont understand!!! ratio and fraction are different!!!!! /s

@potaatobaked7013

26 күн бұрын

We do have to be a bit careful. In your case, the critical mistake is the fact that the limit is as Δx approaches 0, so it shouldn't appear in the result since it was limited to 0 The derivative is defined to be the limit of Δy/Δx as Δx approaches 0. This is the limit of a ratio so in a lot of cases, it ends up behaving as if it were a ratio itself, but there are cases where some properties of the fraction do not transfer to the limit, usually in multivariable contexts. Because it loses many of the properties in a multivariable context, it cannot be considered a fraction. In a single variable context, the properties usually hold however.

@kingbeauregard

26 күн бұрын

@@Endermanv-ot2if Well, fractions and ratios ARE different. A fraction is a piece of a thing, like a slice of pie taken from a whole pie. You could even have 3/2 pies, with the understanding that it's not pies on top and bakers on the bottom. But a ratio compares dissimilar things, for example pies to apples.

@prant55

26 күн бұрын

@@kingbeauregard oh wow thanks, we can also note that we can use ratios as fractions to compare same things; but not fractions as ratios right?

I’ve really wondered about this since taking my first class on differential equations several years ago, and even more since then as this idea has come up in all my physics courses, and this video finally put to rest that query that I’ve never gotten a good explanation for. Thank you so much! This has really helped me a lot.

dy/dx has been lost in the shuffle. dy/dx is limit ( lim ) A->0 of Ay/Ax where Ay = ( y - y1 ) ; Ax = ( x - x1 ). That's it ! Further elaboration: dy/dx may be considered irrational. An irrational quantity is a real quantity which can't be represented as a ratio of two integers ' n ' the numerator and ' d ' the denominator as n/m , m =/= 0 .

Bri saved my Calc grade about 8 years ago...I'm now a ME. Thanks for everything good sir@

That’s some smooth and not-so-subtle transition into and out from the sponsored section, if I‘ve ever seen one!

"It is NOT a Fraction >:( " Chain Rule: Are you sure about that?

@SergioLopez-yu4cu

24 күн бұрын

Proof of chain rule: (f(g(x)))' = Lim ((f(g(x+dx))-f(g(x))/dx). Multiply and divide by g(x+dx) - g(x) and you obtain the rule. Where is the limit treated like a fraction? You are just doing valid manipulations, since g(x+dx) - g(x) ≠ 0 for some dx (g is not a constant function).

@GreyJaguar725

24 күн бұрын

@@SergioLopez-yu4cu I meant eg: dy/dx = dy/dt × dt/dx and that's all it was for me in College, I bet in Uni it goes more rigorous and all that, but I don't know about that. It was also mainly a joke.

@somerandomuserfromootooob

22 күн бұрын

​@@GreyJaguar725, They taught that for me in high school, not any CoMpLiCaTeD.

the biggest discrepancy in the approach from a pure mathematics perspective and everyone else is that a pure mathematician uses the formalized version of calculus (real analysis in one variable) while everyone else is satisfied with the intuition of infinitesimals used in the birth of calculus i'll never forget the time a professor asked a physics student who was nagging ab dx/dy in an analysis course "what is the formal definition of dx and dy in the fraction dx/dy?" and the silence was loud

@dukeofvoid6483

4 күн бұрын

Both dx and dy are indefinitely small variables, but one depends on the other. This is the formal definition.

@trivialqed

4 күн бұрын

@@dukeofvoid6483 thats the case in non standard analysis, proofs for many results are more direct and easier but the avg math curriculum uses standard analysis. in standard analysis one needs many tools to formally define dx/dy: dual spaces, differentials, differential maps, differential forms, integration along C1 curves

@dukeofvoid6483

4 күн бұрын

@@trivialqed 1 = 1, y = y, y + dy = y + dy, y + dy = y + dx(dy/dx) Or to switch from Leibniz to Lagrange: f(x + h) = f(x) + h.f'(x) And note these are all finite variables i.e. we are analysing secants not tangents. If we freely employ algebraic tools and neglect h terms toward the end, we are declaring h to be infinitesimal.

@trivialqed

4 күн бұрын

@@dukeofvoid6483 one needs to define what dx and dy are before being able to do algebraic manipulations involving them. for example if its not define then the statement y + dy = y could imply dy = 0 bc zero is the only element that satisfies the additive identity of a numerical set. but intuitively we know dy is not zero, hence its not an element of the set. then what is dy? in non standard analysis infinitesimals are introduced as elements (hyperreals) of the hyperreal field *R and hence one can easily justify algebraic manipulations with these new elements which intuitively are the infinitesimals the standard approach in a math curriculum is analysis where hyperreals dont exist and hence infinitesimals arent a thing. one constructs dy as a differential form and only until then algebraic manipulations are possible

@dukeofvoid6483

4 күн бұрын

@@trivialqed dx and dy are indefinitely small variables.

I took calculus twice once with a professor who stressed dy/dx IS NOT A FRACTION. The next said you can use it as a fraction the only difference is that dy/dx can be defined for /0 when the equations are applied. I got a C in the first professors class and a B in the second. I felt lost when i couldn’t approach it as a fraction but it all made sense when i did.

@johnjameson6751

22 күн бұрын

dy/dx is a ratio of differential 1-forms on the real line, so you can always treat it as a fraction. However you have to understand that value of dy (or dx) at a point is not a real number, but a linear function of velocity.

@pyropulseIXXI

10 күн бұрын

I got 98% in all my classes and I argued against the professor who said it wasn’t a fraction and I said it clearly was I had to bring up hyperreals and other such notions for ‘rigors’ sake

I would say that the answer to the derivative at a known value IS a fractional value of the instantaneous change in y compared to x. As an example it could be the slope of a function where y depends upon x. So writing it that way actually makes sense. That’s the best I’ve got today. (On the integral side I think most people use the dx notation so it’s a bit symmetric.)

Hearing "anti-derive" at 4:19 instead of "anti-differentiate" hurt my soul.

@jaylenc_

26 күн бұрын

IKR!!! I feel like Americans don't know that derive already has a meaning in mathematics...

@CliffSedge-nu5fv

25 күн бұрын

Sounds like you have a weak soul.

@sugger727p4

17 күн бұрын

@@jaylenc_ Stereotyping a whole country just because a guy with a different accent said something you don't like is crazy

@jaylenc_

16 күн бұрын

@@sugger727p4My point was that it’s not just him… also it’s not really a stereotype, they think they’re right 😅

@sugger727p4

16 күн бұрын

@@jaylenc_ Stereotype: a widely held but fixed and oversimplified image or idea of a particular type of person or thing. In other words, a generalized belief about a whole group. Saying that Americans (the group) don't know X thing (a generalized belief) is a stereotype

Wait until you see us physicists differentiate by a vector lol

I'm reminded of a video "Fixing the Second Derivative (Refactoring Calculus, Part 3)" which discusses problems with notation used for second derivatives.

if it walks like a duck, looks like a duck, and quacks like a duck, its probably a duck

in physics we start with average quantities, like defining average velocity as (delta x/delta t) where both the numerator and denominator are finite changes, so that it IS a fraction, a ratio. Then instantaneous velocity is defined as dividing up to the time interval delta t into, say 100,000 pieces. It's still a fraction/ratio. Then think of dividing it up into 1,000,000 pieces. Still a fraction/ratio. I always tell the class that it just depends on how precise you want to be with the instantaneous velocity. Most real problems are solved on computers, using something as simple as excel, or as complex as a programming language, where you decide yourself how many pieces you want to break the time interval into. In this way of thinking, dx/dt is definitely a fraction/ratio, and will always work.

i really wish i could understand that notation some day. i also heard that the notation is often used with an algebraically incorrect "shorthand", and so now i never know when to trust these expression and how i could restore them to the correct form.

@dukeofvoid6483

4 күн бұрын

Next to nothing - a single paradigm

dy/dx is a ratio of differential 1-forms, so it is a fraction, but at each point, dy and dx are not real numbers, but functions of velocity v: dy(v) says how fast y is changing at the given point when you are moving with velocity v along the real line. Since velocities on the real line have only 1 dimension, a ratio of two functions of velocity (at a given point) is a real number. So treating dy/dx always works in 1-dimensional calculus, as long as you take care of zeros in the denominator, and appreciate that dy and dx themselves are not real-valued functions. By convention, dx is the standard function of velocity, so dx(v) is just v in standard units. In any case dy/dx measures the ratio as to how fast y is changing compared to x.

It is a fraction not only because it works so, but also because it's just differential of y divided by differential of x (dy/dx). So it IS a fraction a-priory

1) Define derivative as a limit of the ratio delta y)/(delta x), as delta x goes to zero. 2) Define differential of a function of x as d(f(x))= f’(x).delta x. 3) Thus, d(x)=1.delta x. Note that (dx=delta x) not equal to 0. 4) We have, dy=f’(x) times delta x= f’(x).dx 5) Now, recover f’(x) by dividing dy by dx. Thank me later.

NO! You don't just take the bottom one off for separable diff eqs..... Bad! It all comes down to the geometry it's describing. df/dx is just a vector. It points in a direction. So, if df/dx is a vector then what is dx? Well dx is an increment on the axis. If df/dx has a length of 5, then it cuts through 5 increment markers of separation dx. The dx in integration acts like our units. If I have something that is 100m long then the length L=100 m. If I change my units from m to cm then the length of my vector component increases to 1000 but my unit/basis gets smaller. My dx is smaller as I'm fitting in 10 times as many dx. It is contravariant. e_i=d_{i}, and e^{i}=grad(x)^i meaning a vector v=u^i d_i = u_i grad(x)^i. I adore LA and tensor calculus, it's a shame that we seem to avoid it at high school but shove calculus down people's throats....

@SergioLopez-yu4cu

24 күн бұрын

You don't know what a vector is if you think every vector has a direction.

@alxjones

10 күн бұрын

@@SergioLopez-yu4cu Every vector does have a direction, depending on what you mean by that word. I would argue that the one-dimensional subspace containing a vector is its direction, so that the projective space of any vector space is the space of its directions.

@SergioLopez-yu4cu

9 күн бұрын

@@alxjones , all vectors can have direction via isomorphism, but they don't really. What's the direction of f(x) = sin(x), which belongs to the vector space of real functions?

@alxjones

9 күн бұрын

@@SergioLopez-yu4cu The direction is [sin] := { r * sin | 0 =/= r in R }.

@DrWhom

7 күн бұрын

for someone who knows tensor calculus, you do talk like a noob

uhh...it is a fraction it is the "change in y over the change in x" as a slope functionality across the entire thing....meaning delta....hence...the "d"....like, I got into a verbal fight with my calculus teacher years ago over this simple shit that was the entire point of the creation of calculus. it's literally the foundation.

The way a Physics student sees it is Operator so you cannot always divide part of the operator tho we do like to do it a lot

@SergioLopez-yu4cu

24 күн бұрын

Well, given a function f:R -> R, the transformation Five: R^R -> R^R such that Five(f(x)) = 5f(x) is also an operator and you can divide by 5 and by f(x) perfectly.

This video is not considering the fact what dy/dx actually is. It is the rate of change of y with respect to x, if y depends on x at all. Triangle y/ triangle x is the average change with respect to x, and it is also a ratio, just like dy/dx, which is the change in y with respect to x when an infinitesimally small amount of change in x is brought, which brings the same small change in y, denoted by dx and dy respectively. Dividing dy by dx gives us the relative change of y with respect to x, that is how y changes when x changes by unimaginably small amounts. In maths, a function's dy/dx means it's slope because its a coincidence that it is the slope, it could've been anything else, but the rate of change of the function is best denoted by the slope at different points in a function.

Great video! Nice to see it worked out the proper way! :) Thanks!

ah yes, dy/dx isnt a fraction, it just is one

Abraham Robinson rigorously proved that using infinitesimals as fractions isn’t an abuse of notation and is mathematically consistent. It’s okay to do it!

Derivative is a fraction. Its definition is limit of (change in y divided by change in x) as change in x goes to 0

the derivative is indeed not a fraction, but it is equivalent to a fraction by definition of a differential. given y=f(x), dy= f'(x) * dx (definition of differentials) this is equivalent to f'(x) = dy/dx. that is a fraction. So yes, a derivative is equivalent to a fraction

Can you make a video on Vn method used in sequence and series

As long as dx or dy does not approach 0 it is actually a fraction. Any student beginning to learn calculus learns this fraction: (f(x0+h) - f(x0)) / h, and what happens when h -> 0. Later you learn some smart methods, such as (f(x) + g(x))' = f'(x) + g'(x), (f(x)*g(x))' = f'(x)*g(x)+f(x)*g'(x), the chain rule, etc. so that later you don't have to consider the fraction you originally learned. These arithmetic rules for differentiation are even derived using the very same fraction. And as far as integration is concerned, dx indicates the width of the individual rectangles in under/oversums, and the integral of a function is found as the limit when dx -> 0, and using the mean value theorem it is shown that ∫f(x) dx from a to b = F(b) - F(a). All using the fraction dy/dx when dx -> 0

That was a great video on the topic!

If dy/dx only is a notation, how can you for example use it in the chain rule? For instance, let us say that a balloon with a spheric shape is being deflated during a time period resulting in a decrease in the volume. On top of that, the radius is diminishing as well. Hence, we get the equation: dV/dt=(dr/dt)*(dV/dr)

@alexdaguy9626

25 күн бұрын

That's the funny part, you can't. Or well, not exactly. The definition of derivative is not a fraction but a limit of a fraction so it makes sense that it may share similar properties to fractions, however it may not be all of them and the ones it does share may not be for the same reasons. I think the chain rule is probably the best example of why you can't really treat it like a fraction. There is a common erroneous proof of the chain rule using the limit definition that is based precisely on the idea of cancelling the fraction where dy/dt * dt/dx = dy/dx. In fact it looks basically the same except you actually write out the value of the dy's and whatnot for a small change in x. However it does not consider the fact that if you do that, the dt could equal 0, and you can't divide by 0, so the left hand side if treated as a fraction is invalid. In fact it could equal 0 for arbitrarily many values of x arbitrarily close to where you are trying to differentiate at which makes things really complicated. There are many ways to remedy this proof, the cleanest one I've seen is to define a function that fills in that gap and show it's equivalent to the original limit. But the point is that the chain rule is non-trivial to prove, it is not quite as simple as cancelling a fraction. It's nice that we eventually have this result, that makes the derivative act very similarly to fractions. However, as with many nice results in math, we should not take this for granted and just call the derivative a fraction, as there are steps going on behind the scenes that make it work.

@SergioLopez-yu4cu

24 күн бұрын

The chain rule written like this: df/dx = df/dy * dy/dx is an abuse of notation too. Chain rule can be proven just using the rigurous definition of a derivative.

If not a fraction, why fraction shaped?

It is a fraction because it is illegal to divide by 0. That’s why we are allowed to treat it as a fraction. It’s probably one of the most important concepts to understand.

@rickperez8975

22 күн бұрын

Remember it’s the limit as h approaches 0, and it is “infinitesimally small” but not really zero because we cannot divide by zero. Once you start doing numerical analysis maybe through coding, you see that dx and dy are real, so therefore dy/dx is real as well. dy/0 cannot exist. But dx is not 0, it’s just really small.

@DrWhom

7 күн бұрын

@@rickperez8975 you are talking nonsense

@rickperez8975

7 күн бұрын

@@DrWhom you just don’t see it

@rickperez8975

7 күн бұрын

@@DrWhom the way I saw it was when I was programming PID controllers and was using the differential component. It made most of the questions I had click. Then the theory kind of falls into place. I.e; Can’t divide by zero, but if the dx component is unfathomably small, then technically you can divide by it, more so because dx (step size) is constant for all dys

If we think of our curve as a curve parameterized by t, and we think of dx as being x'(t) and dy as being y'(t), then we can think of dy/dx as being a fraction (so long as dx is not zero).

It's a fractions if we understand dy, dx as differential forms of degree 1. Linear in tangent vector....

Sponsor ends at 3:40

Could you make a video on exotic integrals using different metric spaces ie: lebesque integrals?

Sorry, as a physicist I have to plug my ears and go lalalallala to any idea that I can’t treat it as a fraction..

If you treat them as 1-forms on a 1-dimensional manifold, then they're indeed fractions.

Where can I watch the video showing the integral of x raised to dx minus 1 (as shown on 7:11 of this video)?

It works because the two operators are equivalent; that is dy/dx dx is equivalent to dy. although you can think of it has a fraction taken to its limit has delta x approaches zero!

Thank you very much. This is an eyeopener, indeed.

So we can just assert that the fractional notation dy/dx is simply a specifc case under certain parameters of the general and all knowing form of y prime. The fractional notation helps in algebraic manipulation but only under certain cases, which is good enough regarding the application of the derivative; in physics, engineering and so on. But to be mathematically correct in all cases, we need to acknowledge the shortcomings and limits of the fractional notation.

Treating dy/dx as a fraction is how I got through diff eq

The thing I don’t get is the sense in which this isn't a fraction... if you do calculus from first principles you get the limit of a specific fraction. It is the rate of change of one variable with reapect to another which is a ratio ir in other words a fraction. The only sense in which it doesn't make sense as a fraction is that dx doesn’t mean anything useful on its own, as essentially dy/dx is a specific well defined verion of 0/0.

change in y, over change in x It's literally a fraction It's literally the slope of a line

I agree with your opinion. I like dy/dx becuase it makes it clear which variable you differentiate with respect to. Although I did right a lot of prime notation when I was a student. 🤷‍♂️

It only works for the one dimensional examples. If x is multi dimensional, then dy/dx and dx do not cancel like that. But the nice things is that you can replace multiplication by dot product and it works again. This all boils down to the chain rule. Yes, mathematically, they are not some objects that can cancel out. But the notation is chosen cleverly and reflects how the chain rule acts.

@DrWhom

7 күн бұрын

wrong

I’ve been wondering how this works for so long

Well, you shouldn't start using it as a fraction. That's where a lot of teachers get it wrong.

I beg to differ on the title assertion! The caveat that "dy/dx isn't a fraction, it's notation" is a holdover from the old definition of differentials as "infinitesimals". The modern definition of differentials is as (perfectly finite and reasonably valued) "local linear coordinates" specifically for points on the tangent line relative to an origin at (x,y) on the curve. Thus dy/dx is the ratio of those coordinates given dy = m dx for m = f'(x), that being the slope of the tangent line at (x,y) on the curve y=f(x). In my calculus class I teach differentials first as limits of difference quotients (Gateaux differentials) and then the derivative is, by definition the operator mapping differentials to differentials and in single variable calculus that is their ratio! BTW: fractions are also "just notation" :)

I believe that it works because it IS a fraction. I see it as the rate at which y is increasing with respect to the increasing rate of x at the infinitesimal level. (Please note that "rate of increase, can also be negative - in which case the variables are decreasing at the infinitesimal level). But what do I know?

Thank you. 😊

I always think dy/dx IS a fraction. It is the limit of a fraction of f(x+h)-f(x) and h, when h approach 0

@uKaigo

22 күн бұрын

That's my thought too

@peterfireflylund

21 күн бұрын

Yes, but why should it make sense to take something OUT from inside the limit operation and treat it like a normal variable?

@user-ej7sr3ow8b

20 күн бұрын

@@peterfireflylund For this question, I'm actually more interested in why taking limit should make the fraction not an fraction anymore?

@BlueGiant69202

10 күн бұрын

One can also define derivatives as the ratio of two directed integrals. Geometric Calculus 4 by Alan MacDonald kzread.info/dash/bejne/nWZhm7Kppc7NfaQ.htmlsi=fgRHlCfRpCt8LSTa

Every time I see a person calling 'dx' an infinitesimal a part of my soul dies

This is why calculus classes shouldn't skip the differentials chapter. Understanding that dy = (dy/dx)dx as a matter of definition & notation (rather than algebra) is covered in those chapters, as an easy consequence of the chain rule.

But it literally is a fraction in the hyperreal number system

I’m saying it dy/dx is a fraction The Dirac delta function is a function And sinx = x

I wish you gave an example where it does not work as a fraction, I'm curious what kind of cases I should be wary of when treating derivatives as fractions.

@viliml2763

18 күн бұрын

With single-variable derivatives, there are no such cases. You can safely treat it as a fraction all the time, it has been rigorously proven that it works. With partial derivatives, you can run into problems.

@JESUS_CHRlST

18 күн бұрын

Either partial derivatives or higher derivatives are where the problems come in. You might have to scroll a bit but see the comment I made, basically you’ll be fine as long as it’s the first derivative and not partial

@PeachCrusher69

10 күн бұрын

​@@JESUS_CHRlSTexample for higher derivatives giving a problem when treated as a fraction?

the chain rule itself is derived by treating it as a fraction. In fact, it is a fraction because it refers to slope which is small change in y divided by small change in x

@carultch

25 күн бұрын

That's just a shortcut. There are limit proofs of the chain rule that are more bulletproof than proofs involving treating the notation as a fraction.

@SergioLopez-yu4cu

24 күн бұрын

No, it isn't

I remember learning about differential equations in College. I went to several of my math (and physics) professors asking them "hey, what's the nature of these differentials? Since 𝑑𝑦/𝑑𝑥 isn't a fraction, it doesn't make sense to me that 𝑑𝑦 and 𝑑𝑥 are numbers. Are they some sort of vectors? Do they exist within ℝ?" and I universally and independently got the same answer from several different people, each with different academic backgrounds. "Yes, the differentials are real numbers, and yes, 𝑑𝑦/𝑑𝑥 is a fraction". Good enough for me

It is a fraction! It is the limiting fraction when both f(x) and x tends to zero.

Was nearly a full half of the video an advert for Brilliant? That's not a fraction I like.

Except it IS a fraction. It's the ratio of the change in the dependent variable to the change in the independent variable

@f-zg2yw

5 күн бұрын

exactly how ive always treated it, its self implied in the definition of the limit, as delta x tends to zero delta x tends to dx and thus the corresponding delta y becomes infinitely small and tends to dy....

@dukeofvoid6483

4 күн бұрын

Other way round...

@freshmozzarello

Күн бұрын

@@dukeofvoid6483 oh shit you're right

Just treat as a fraction, then subs dx=0 after completed division. d/dx is an operator that is consistent with the operation described. QED.

it's a genuine fraction, the ratio of two differentials

It is indeed a fraction often interpreted as rate of change of y wrt x.

From an engineer’s point of view, dy/dx is delta-y/delta-x as delta-x approaches 0 or (y2-y1)/(x2-x1) as x2 approaches x1. I’m not sure what I’m proving here but the latter expression can legitimately be multiplied by (x2-x1) which then cancels. It seems logical to me lol.

Don't you think chain rule also treats derivative as fraction

It’s a fraction just fine, just also a limit.

So sad I wasn't shown this in Calculus!

@reh3884

24 күн бұрын

Why? He just making it harder than it has to be.

@tommyhuffman7499

24 күн бұрын

@@reh3884 In the early Calculus days it felt like they were just making up the algebraic rules for what could and could not be done with Calculus. For anyone who thinks deeply about what is happening during each step of a problem, blindly copying an example that doesn't make sense is the harder way.

But, it is a fraction? It is the ratio between an infinitesimally small difference in x vs an infinitesimallly small difference in y. we're dividing it.

this is all nice and cool, but in vector calculus or manifolds students get introduced to forms and dy/dx is indeed a fraction of two differential forms

this only happens because the standard notation is not quite right. don't use differentiation. the d[] operator is an implicit diff. you can really see where it went wrong by implicitly differentiating 1/dx. See Johnathan Bartlett's notation change. d[c]=0 means "c is constant" d[d[t]]=0 = d^2[t] means "c is a line" d[a + b] = d[a] + d[b] d[a * b] = d[a]*b + a*d[b] d[a^b] = b a^{b-1} d[a] + log_e[a] a^b d[b] d[log_a[b]] = ... complicated, but derivable from d[a^b] The second derivative is where the standard notation goes wrong. This is the actual second derivative. Apply d first, only divide by dx as a separate step. d[ d[y] / d[x] ]/dx = d[ dy * dx*{-1} ]/dx = ( d[dy]/dx + dy*(-dx^{-2}*d[d[x]]) )/dx = d^2y/(dx^2) - (dy/dx)(d^2x/(dx^2)) The thing that most people wont calculate right is d[ 1/dx ]. When people think of acceleration, they assume that d^2x = 0. This is true when x is a line, when the variable is t, for instance. The third derivative is even more complicated. But you can check this for second derivative and see that you can use it to solve for dy/dx. That subtracted term is usually zero, but you need to keep it around for the algebra to work. z = [ x^2 + y^2 ] dz = 2x dx + 2y dy One partial is to set dy=0 and dx*dx=0 and dx>0 dz/dx = 2x dx/dx + 2y dy/dx

Thing is, the chain rule itself is most easily justifiable by thinking of the derivative like a fraction: like, if I have dy/dx = y'(x) => dy = y'(x) dx Suppose now that dx/dt = x'(t) Meaning that: dx = x'(t) dt Therefore dy = y'(x) * x'(t) dt => dy/dt = y'(x) x'(t) Notice that at more or less every step I'm manipulating differential elements of x and t like they're fractions? I don't think it's sufficient to say that fractional behaviour can be satisfactorily explained with the chain-rule, because it's not clear that whatever's behind the chain-rule isn't itself just treating derivatives like fractions.

dy/dx = y-y_a / x-x_a so it's a fraction >

i learnt calculus mainly from 3b1b and i had no problem because he actually derived it to be a fraction. i dont know why people think it isnt

@SergioLopez-yu4cu

24 күн бұрын

Because it's not defined that way and treating it like a fraction when it isn't is something an engineer would do.

@viliml2763

18 күн бұрын

@@SergioLopez-yu4cu If it looks like a fraction, walks like a fraction, quacks like a fraction, then it's a fraction. Mathematicians of all people should look at things abstractly in terms of their properties.

@SergioLopez-yu4cu

18 күн бұрын

@@viliml2763 , formalize "looking like a fraction" (which is pure notation), "quacking like a fraction"... Real analysis is constructed with rigurous definitions, modern maths try respect the definitions and there's no problem with that because you can prove everything you need treating derivatives with the definition.

But, isn’t the chain rule a consequence of treating dy/dx as a fraction?

Why does it works in differential equation ?