So Why Do We Treat It That Way?

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[@BriTheMathGuy]

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[@Frankiethedogishere]

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

[@voidisyinyangvoidisyinyang885]

can you do a noncommutative nonlocality math vid a la Basil J. Hiley or Alain Connes? thanks

[@vivada2667]

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

[@user-iv4dh7zp7s]

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

[@cubicinfinity2]

@@user-iv4dh7zp7s lol yes.

[@davidbrown8763]

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]

It doesn't work in case of double derivatives

[@vignotum132]

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

[@naytte9286]

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]

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]

Please... that book is barely anything ;)

[@martimzurita]

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

[@_ranko]

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]

​@@_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.

[@ManilockPi]

If not a fraction, why fraction shaped?

[@user617a6d6]

it's a ratio. small change in y over small change in x

[@tka3]

​@@user617a6d6 A ratio can still be expressed as a/b so its still a fraction in a sense

[@alessandrocoopman9135]

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

[@kacpermalinowski8215]

Physicists will also do this, so idk if that's a good argument kzbin.info/www/bejne/hKuuYmR6p7eXodUsi=uexzXJEVOzibOdco

[@ohlala2219]

Yeah they always do these sort of things to math xD

[@intellix7133]

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

[@juanmanuelmunozhernandez7032]

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]

@@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?

[@eliteteamkiller319]

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]

the what is WHAT

[@themathematicstutor4092]

Taking the d with respect to x

[@nestorv7627]

I love phallic math

[@frederf3227]

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

[@w花b]

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

[@jmcsquared18]

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]

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]

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.

[@lschlickmkw]

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

[@alb2451]

@@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]

@@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.

[@ingGS]

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

[@johnjameson6751]

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

[@v12-s65]

@@johnjameson6751for small approximations mostly 😜

[@SoloRenegade]

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

[@ingGS]

@@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]

@@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.

[@slava6105]

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).

[@mikey-hm7dt]

Thats gotta be the most anti-algorithm title ever

[@SeeTv.]

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]

I was just thinking that 😂

[@uKaigo]

​@@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]

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]

@@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)

[@matthewmason2538]

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]

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]

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]

@@JktuJQ 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]

it doesn’t work in multivariable calculus

[@lukandrate9866]

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

[@maringenov7753]

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]

exactly - basic stuff, right?

[@maringenov7753]

@@DrWhom Right :)

[@live_free_or_perish]

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

[@professorpoke]

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

[@f_add_mebowshot5677]

Wise words

[@sieni221]

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]

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.

[@Thesaddestmomentinourlives]

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]

so you're saying it's a fraction

[@Endermanv-ot2if]

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

[@potaatobaked7013]

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]

@@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]

@@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?

[@smesui1799]

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 .

[@v.r.kildaire4063]

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]

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]

​@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]

@@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]

​@@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]

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

[@OBGynKenobi]

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

[@stumbling]

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.)

[@TomKorner-ye4lf]

Bro really just proved the entirety of physics, my my

[@GreyJaguar725]

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

[@SergioLopez-yu4cu]

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]

@@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]

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

[@Guywiththetypewriter]

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]

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]

I like how you write

[@DrWhom]

spoken like a true engineer 100% confused bullcrap

[@aspartamexylitol]

​@@DrWhomcope

[@solidpython4964]

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.

[@marscience7819]

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.

[@kevin-gg8ir]

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

[@evilotis01]

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

[@kurtrosenthal6313]

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]

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.

[@martinsanchez-hw4fi]

There is a case that I think is more subtle and it is on manipulation of the differentials inside an integral, not even substitution. Like in the work energy theorem, where you go from m\int_{x_1}^{x_1} \frac{dv}{dt}\ ,dx to m\int_{x_1}^{x_2}\frac{dx}{dt}\,dv

[@Oklahoma-Dreaming]

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.)

[@TheSuperiorQuickscoper]

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

[@jaylenc_]

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

[@CliffSedge-nu5fv]

Sounds like you have a weak soul.

[@sugger727p4]

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

[@jaylenc_]

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

[@sugger727p4]

@@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

[@thea.igamer3958]

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.

[@johnjameson6751]

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.

[@Eichro]

*How* is dy/dx not a fraction? Is a derivative not defined as lim (x2 -> x1+) (f(x2) - f(x1)) / (x2 - x1)) ? Is dy not the numerator and dx the denominator?

[@alexanderehrentraut4493]

As you said, dy/dx is the limit of a fraction. And while a fraction is a quantity divided by another quantity where you can separate these quantities and still get a true result, the same is not generally true for the limit of a fraction, which is a singular quantity. In the case of the derivative, splitting this limit would be to divide the limit of the numerator by the limit of the denominator, which would yield 0/0.

[@Eichro]

The thing is, a limit towards 0 is different from 0. It's not absolute 0 / absolute 0, it's too-tiny-to-yield / too-tiny-to-yield, and they'd have an actual, sensible ratio between them.

[@Uejji]

dy/dx notation for derivatives is fundamentally related to delta-epsilon notation for limits. A very small change in x results in a very small change in y in a way that is consistent throughout the entire function. dx and dy can be thought of as actual numbers that make the equation true. There are infinitely many of them, but they have a defined functional ratio, expressed as the derivative of the function. However, because they are actual numbers, they can be operated on as actual numbers. eg f(x)dx + g(x)dx = (f(x) + g(x))dx etc

[@thewok3576]

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]

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]

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.

[@pyrotas]

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

[@sofia.eris.bauhaus]

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.

[@eliasrodriues6614]

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

[@mkrsinfo2859]

dy/dx is literally as described as a quantity reduced to the smallest possible value(the limit) in fractions we have the freedom to set the part of a quantity where a quantity differentiated is the smallest value reduced and when it is integrated it is simply the joining of continuous bodies

[@Hk_498]

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.

[@pengutiny6464]

x + h - x = dx [horizontal] f(x + h) - f(x) = d f(x) [vertical] d/dx f(x) The ratio stay the same, since we scale the distances equally

[@davidbrown8763]

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?

[@jespermikkelsen7553]

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

[@user-lz1yb6qk3f]

It really is a fraction, but it's under limit/standard part (depends if you derive calculus from limits or from hyperreals). You can derive all the algebraic properties of the fraction using properties of limits/standard part. You basically can treat it like a fraction.

[@VaraNiN]

Wait until you see us physicists differentiate by a vector lol

[@starhacker6411]

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

[@trimsclapped2667]

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

[@kevskevs]

Shouldn't the "d"s in the differential operator be written upright rather than italic (from a LaTeX point of view)?

[@hgp314]

no

[@JannikGanswindt]

Yes, they should.

[@slava6105]

Also LaTeX's recommendation (or mathjax, idk) is to add a special space inbetween the subintegrable expression and the differential if it is written separetely from others (not embed like ∫(dx/x), separately would be ∫(1/x)dx with soace before dx)

[@sieni221]

Stylistic choice. Most people write differential in italics some not.

[@gerardvanwilgen9917]

I think so, and this indeed seems to be the norm in most non-English-speaking countries. There π, e and i are also often printed in Roman type, which makes sense because there are not variables but constants.

[@KelfranGt]

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]

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]

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

[@trivialqed]

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]

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

[@trivialqed]

@@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]

@@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]

@@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]

@@trivialqed dx and dy are indefinitely small variables.

[@isooo8175]

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

[@sabriath]

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.

[@rickperez8975]

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]

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]

@@rickperez8975 you are talking nonsense

[@rickperez8975]

@@DrWhom you just don’t see it

[@rickperez8975]

@@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

[@sphinx1239]

The term 'dx' indicates a small change in x (or that x approaches zero), and the term 'dy' indicates the small change in y caused by that dx. Now the derivative is just the small change in y divided by the small change in x, dy/dx.

[@Laicicles]

Sponsor ends at 3:40

[@J_CtheEngineer]

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. 🤷‍♂️

[@tlpiwsool2d]

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

[@roderictaylor]

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).

[@berryesseen]

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]

wrong

[@joshuaasuman1]

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

[@amosdominguez3685]

Or you can use the definition of the differential... dy = (dy/dx) dx... and substitute dy in the integral

[@ikarienator]

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

[@ShanBojack]

I might be wrong but i read somewhere that dy/dx can be treated as a fraction no problem but not the higher powers/derivatives i.e. for e.g. d²y/dx² or d³y/dx³ can't be treated as fractions/ratios?! Please tell me someone

[@slava6105]

d²y/dx² = d(dy/dx)/dx or (d/dx)(dy/dx) (this is operator application btw, not _just_ parentgeses to mark multiplication, but it is really seamless this way)

[@ShanBojack]

@@slava6105 i understood what it meant but it still doesn't answer my question 😥

[@slava6105]

@@ShanBojack I generally have small knowledge on this topic (differentials are hard). First order ones work perfectly but others are not if i remember correctly. P.S. while writing it, found some post on reddit stating this is indeed just the notation but it works because and because Leibniz with Newton defined it as small numbers but we do not (or do we, i don't understand most if it anyway).

[@Tzizenorec]

Where is d²y/dx² from? At first glance I suspect that's a broken notation. The variables shouldn't work out that way. Let's see... Starting with y=x²: dy = 2xdx d²y = 2dx²+2xd²x Oh, I see now. d²y/dx² is actually a partial derivative of dy with respect to x. It implicitly assumes that d²x=0.

[@Tzizenorec]

I tend to think the professors who teach calculus this way are just... teaching low-level wrong stuff. Like when your teacher in kindergarten told you that you can't subtract a bigger number from a smaller number. dy/dx works only if x is the _only_ variable in y, and it works because if x is the only variable then every single term will wind up multiplied by dx. So in this case, "dy is a multiple of dx" is literally true. If there are multiple variables involved, then you'll wind up with some terms multiplied by the derivative of each of those variables. So, say, if y=x^2+t, then dy=2xdx+dt. You _can_ divide that by dx, winding up with dy/dx=2x+dt/dx; but most math professors like to discard the dt/dx term. This is where the "partials" idea comes from, and it isn't really valid math but it works if you either reassemble the partials into the full equation later, or set the derivatives of some of the variables equal to zero. The second derivative of y has this same problem, because the derivative of a function in x is almost always a function in _two_ variables: x and dx. (Sometimes the x disappears leaving only dx, but not usually.) So, to do it actually rigorously correct, treat dy/dx as division and do _not_ perform the "d/dx" operation that implicitly discards the dx. You might accidentally wind up doing a partial derivative and throwing away variables that you shouldn't have thrown away. (I had the good fortune to have learned and absorbed the concept of "dx" as an independent variable, and dabbled a bit in differential equations, before I ran into a calculus professor who wanted to teach me about partial derivatives.)

[@abhirupkundu2778]

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.

[@someknave]

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.

[@Peter_1986]

Isn't dy/dx basically just an infinitely precise version of the slope formula Δy/Δx?

[@teddy_0932]

Thank you so much for making this video Bri! I major in physics and I put a lot of emphasis on the strict structural proof of math. That’s why it baffled me so much when my teachers treated derivatives as fractions when I was a freshman learning calculus. My classmates just went along with it but this question stuck with me for years. I’m about to enter my fourth year in university and eventually I just accepted it. It wasn’t until this video that I got my head around the whole concept. Thanks again!

[@captainspirou]

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!

[@EvilDudeLOL]

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

[@aadik4458]

But (sorry if this is a dumb question) doesn't the chain rule also imply dy/dx is a fraction? Atleast it looks like it does to me Please correct me

[@aroundandround]

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

[@user-ej7sr3ow8b]

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]

That's my thought too

[@peterfireflylund]

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]

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

[@BlueGiant69202]

One can also define derivatives as the ratio of two directed integrals. Geometric Calculus 4 by Alan MacDonald kzbin.info/www/bejne/oWOQaoOmqtCag6Msi=fgRHlCfRpCt8LSTa

[@The_Green_Man_OAP]

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.

[@SoloRenegade]

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

[@advocatusdiaboli7851]

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

[@mechablade4756]

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

[@jamesbaugh8001]

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" :)

[@JESUS_CHRlST]

To repeat my reply to another comment, the problem of not being a fraction is not at all just in partial derivatives. Any derivative higher than the first in dy/dx = 1/(dx/dy) is untrue. Try differentiating this yourself to see why You will find d^2x/dy^2 = -f’’(x)/(f’(x))^3 Where f’(x) = dy/dx As you can see there is a differential relationship between a second derivative (or higher) and its reciprocal, so it is not a fraction Still shocks a lot of people

[@evilotis01]

thank u jesus

[@JESUS_CHRlST]

@@evilotis01 got u bro

[@snipesjuanr7762]

Well that's not surprising, cuz d²y/dx² is not even same as d²x/dy²

[@viliml2763]

You have no idea what you're talking about. Your example doesn't disprove the fraction view of derivatives. Here, watch me prove the inverse second derivative formula using fractions: dy = a dx (dy/dx=a) d(dy/dx) = b dx (d^2y/dx^2=b) d(dx/dy) = d(1 / (dy / dx)) = - d(dy / dx) / (dy / dx)^2 = - b dx / a^2 = -b/a^3 dy (d^2x/dy^2=-b/a^3)

[@JESUS_CHRlST]

⁠​⁠​⁠@@viliml2763That is a correct result and the same as in my original comment. Except that that does disprove it being a fraction for the second derivative, as the final result is not 1/b. As you’ve proved there in full, the reciprocal of the second derivative of y with respect to x is not the second derivative of x with respect to y. So it’s not a fraction for the 2nd derivative

[@johnholme783]

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!

[@nogodnowhy2629]

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

[@dominicellis1867]

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

[@genessab]

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..

[@joep_s4878]

I’ve been wondering how this works for so long

[@tommyhuffman7499]

So sad I wasn't shown this in Calculus!

[@reh3884]

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

[@tommyhuffman7499]

@@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.

[@robfielding8566]

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

[@romh7261]

I always thought of dy/dx as a slope?

[@GaSevilha]

isnt it a ratio of infinitesimals same as dy*dy is the area of infinitesimals which is the whole reason why summing it all gives you the total area of the curve?

[@freshmozzarello]

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]

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]

Other way round...

[@freshmozzarello]

@@dukeofvoid6483 oh shit you're right

[@skilz8098]

You mentioned that dy/dx is not a fraction, yet we can treat it like one through the properties and applications of the chain rule through the fundamental theorem of calculus. I'm not disagreeing with this. Yet, there is another perspective to consider as to why this works. You also stated that dy/dx is a notation and that we use it as an operator. This still holds true. What is this other perspective? It is based on the actual definition of the derivative of a given expression, equation or function. The basic definition of the derivative is finding the slope or gradient of a given curve where the slope of the curve at that given point is a point of tangency. This leads us back to a combination of both algebraic and geometric notations and forms of slope or gradient which is the essential part of linear equations, expressions and algebra. We can find the slope of any linear equation of the form y = mx+b using the slope-intercept form of the line. Here b is the y-intercept and for this demonstration we are going to set this to 0 so that the y-intercept is at the origin (0,0). The slope m is defined as rise over run which can be solved for any two points on a given line, line segment or vector from m = (y2-y1)/(x2-x1). This is basic algebra. We can substitute y2-y1 with deltaY and x2-x1 with deltaX which will simplify the equation to m = deltaY/deltaX. This slope is the ratio of the differences in the change of both orthogonal directions. When we move on from basic Algebra and finding slopes of straight lines, and beyond that of understanding the polynomials, and other various algebraic functions is when we begin to get into calculus. However, we typically learn trigonometry first since many of the methods of finding derivatives and integrals requires us to know and use many of the properties and functions within trigonometry. Without first knowing it or realizing it there is a direct relationship between Algebra, Geometry and Trigonometry. There is a direct relationship between linear equations, the polynomials, radicals, logarithms, and the trigonometric functions. Regardless of the dimensionality of the curve or function that we are trying to find its derivative, in essence we are trying to solve for its slope at the point of tangency and this slope is a gradient linear expression or a ratio of. Here's the connection between these fields: We can take the expression y = x. Here the y-intercept b = 0 and the slope m = 1. This is the simplest of all linear equations. If we put in 3, we get out 3, if we put in 9, we get out 9, and so on, and this is true for all Real & Complex numbers however, we will only be focusing on the set of all Real Numbers. This line is bisector of the XY plane. We also know that the coordinate axes of the XY plane denoted as (Xi, Yj) where (i,j) are component unit vectors are orthogonal or perpendicular to each other. They have a separation or rotation of either 90 degrees or PI/2 radians. With that, we know that the slope of y = x is 1 and we also know that the angle of this line between both the Y and X axes is 45 degrees or PI/4 radians since it is simply 45 degrees / 2, or PI/2 radians / 2. We know that the slope formula is m = deltaY/deltaX. This ratio of difference without knowing it is also the tangent of the angle between the line of the form y = mx+b; b = 0 and the +x-axis. A simple one statement proof: tan(45) or tan(PI/4) = 1. This is always true. Now to prove that m = deltaY/deltaX = tan(t) we can use the trig identity tan(t) = sin(t)/cos(t). We can test this by checking to see if both of the following are true: deltaY = sin(t) and deltaX = cos(t). We can take the linear equation using the slope-intercept form y = mx+b with b = 0 and substitute these in without changing the operation or transformation of the expression. These are equivalent. y = mx = tan(t)x = sin(t)/cos(t)x (continued...)

[@skilz8098]

(...continued) When we get into calculus there is one additional thing that separates it from elementary mathematics and that is the application of limits, yet within this demonstration the concepts of limits are not of importance. What is of importance is how we derive the equation to find the derivative. At first, we start off with finding the slope of a secant line at a point (a, f(a)) to estimate the rate of change. We can find the slope of this secant line by choosing a value of x near a and draw a line through the points (a,f(a)) and (x,f(x)). This is the difference quotient: m[sec] = (f(x) - f(a)) / (x-a) We can also calculate the slope of a secant line to a function at a value by using this equation and replacing x with a + h, where h is a value close to a. With this we can find the slope through the points (a, f(a)) and (a + h, f(a+h)). This then gives us: m[sec] = (f(a +h) - f(a)) / a + h - a == (f(a + h) - f(a) / h This is the difference quotient with increment h. This is a good approximation for the slope or rate of change, but this only gives us an estimate or "average" rate of change so to speak. What we are looking for is a precise or instantaneous rate of change. For this reason, the secant line will not suffice. In order to solve for this dilemma this is the only place that the limit truly comes into play as we begin to shrink or narrow down these two points of the secant line closer and closer together where the length of this secant line approaches zero and converges to what we call the line of tangency or the point of tangency. The above them becomes: m[tan] = lim x->a (f(x) - f(a) / (x-a) provided the limit exists. Then we can define the tangent line to f(x) at (a) to be the line passing through the point (a, f(a)) having slope: m[ tan] = lim h->0 (f(a+h) - f(a)) / h provided the limit exists. The process of conversion from using the secant line to find the tangent line or line of tangency is directly connected with the slope formula (y2-y1)/(x2-x1) = deltaY/deltaX. This is why deltaY/deltaX = tan(t) where t is the angle between the line y = mx and the +a-axis. Without this linear relationship of the slope and the trigonometric functions, our current notation and understanding of derivatives and antiderivatives or integrals would be impossible. Understanding these connections is another perspective as why we can treat the notation or operate dy/dx as a fraction. Because in truth this operation or the result of the operation itself is not a fraction, yet the operator is a fractional operator in that, it is based on the ratio proportion, difference of, rate of change between the directions in X and Y respectively. Yes, it is an operator and not a fraction by definition, yet we can treat it as if it is one not because of the chain rule, or even through the fundamental theorem of calculus. The chain rule is a result of these relationships, and the fundamental theorem of calculus is built on top of these rudimentary properties that are founded in linear algebra, systems of linear equations, vector notation, matrix applications, and all of the linear transformations starting with translation a horizontal displacement. So yes, we can treat dy/dx similar to deltaY/deltaX , sin(t)/cos(t), and tan(t) since they are all related. Different notations are used within different contexts, yet they are completely substitutable with each other. Their definitions are different, yet their properties are the same. And since a fraction or a ratio by definition is division and if we look at divisions in terms of repeated subtraction as opposed to the inverse of multiplication, this will make a lot more sense. We are repeatedly subtracting small incremental amounts to narrow in on a specific position, location, value. This is why I consider m{rise/run}, deltaY/deltaX, sin(t)/cos(t), tan(t), dy/dx to all be equivalent classes. At the end of the day taking the derivative isn't directly finding the slope as it may yield a function or family of functions, but if we apply the derivative at that point, we are in truth looking for its slope which is in fact a ratio proportion, a fraction. (continued...)

[@skilz8098]

(...continued) There's more to it than just this, such as the Pythagorean Theorem and the definition or Equation of the Circle are directly related and how the simple arithmetic operation of 1+1 = 2 is the unit circle located at the point (1,0). This is why the trigonometric functions have a Pythagorean Identity. This is all possible simply because we are able to add, count, enumerate, to translate a position or a value to a new position or to have a new value. In other words, there's two and only two states to any given operator. In the act or attempt of trying to apply the operator, was the process of the operation generative? If yes, then a translation or transformation took place. If no, then it was a no-operation. Consider these two expressions: 1 + 0 = 1 0 + 1 = 1 We know that basic addition is commutative, associative, etc... Here the two equations above are equivalent. Both the commutative and associative properties define this. Adding 0 to a given value does not change that value, this is associative. The order of the operands in relation to their operator does not matter. So, 1 + 0 = 0 + 1. To better understand this, we can closely examine the position of the operands in relation to their operator. By convention, this will be from Left to Right with LHS and RHS to designate the position of the operand and OP() to designate the operator. We know that both of these equations give the same ending result. Yet if we were to think of these expressions in these terms: LHS OP() RHS Where the LHS operand is the initial or starting position or value, we can then ask the following: Does the act of Applying OP() by RHS change or effect LHS? In other words, was the operator generative? Now we can apply this to the above equations: Does the act of applying the OP(+), addition of RHS(0) onto LHS(1) change the value or state of LHS? 1 + 0 = 1. No, 0 does not change one, therefore the act of applying the OP(+) is not generative, it is a NOP (No - Op) borrowed from terminology in computer science. Likewise: Does the act of applying the OP(+), addition of RHS(1) onto LHS(1) change the value or state of LHS? 0 + 1 = 1. Yes, 1 does in fact change the initial state or value of 0. Here the OP(+) is generative, a translation or transformation has occurred. Without even knowing, the addition operator in itself is a binary system. Now the result of the addition the order doesn't matter, they both end up with the same result. How they got there, is different. In the first example, the operation was a no-op, and in the second example, a transformation or translation has occurred causing the act of applying the operator to be a generative one. So, for every operator OP(), the act of applying it is either generative or a no-op. These are some of the reasons why I love math. There is a relationship between all of it even when you think that there isn't one. Now, for a more challenging question: how do go from 0 to 1... is another topic in itself.

[@kelly4187]

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

[@duncanrichardson2167]

Whether or not dy/ dx etc are fractions it cannot be good practice to assume that the dz s in dy/dx = dy/z . dz/dx can be cancelled. This is because there is no assurance that y as a function of z is the same function as z is a function of x.

[@kabootarkhanawala8271]

It’s a section of the cotangent bundle and should be treated as such

[@RostislavArts]

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

[@thebeardman7533]

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]

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.

[@betaorionis2164]

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

[@DrWhom]

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

[@tulliusagrippa5752]

It may not be a fraction, but it sure behaves like one.

[@mosespeters5546]

Doesnt the chain rule itself treat derivatives as fractions? Like if dy/dx = dy/du x du/dx, the du's cancel out and thats why it works. Thats how i interpreted it, what am i missing here?

[@carultch]

That's not how the chain rule is formally proven. There are limit proofs of the chain rule that are much more rigorous than treating the notation as a fraction.

[@reh3884]

@@carultch That's doesn't change the fact it works as above.

[@SergioLopez-yu4cu]

​@@reh3884, of course, it works, but that doesn't mean you can treat like a fraction every derivative just because the composition of functions verifies that (which is pure luck, since the theorem is proven with a limit).

[@pedroteran5885]

​@@reh3884Well, your argument is: -If my bicycle were a car then it would have wheels. -My bicycle has wheels. -Therefore my bicycle is a car. That is: -If dy/dx were a quotient then the chain rule dy/dx = dy/dt • dt/dx would be correct. -The chain rule is correct. -Therefore dy/dx is a quotient.

[@baptistebauer99]

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

[@ac30428]

It’s a fraction with a limit in front