Just leaving a comment to address some common objections. Yes, this is the limiting case of the quantum derivative (mentioned at the end). No, it does not extend well to higher dimensions. Yes, the multiplicative definition is undefined at zero. There is a fix for this mentioned at the end, f'(0) = lim x--> 0 of f'(x), if this limit exists. But an an important caveat that I did overlook, is that even this 'patch' only works for functions that are continuously differentiable on their domain - an exception would be something like f(x) = x^2 sin(1/x). More importantly, I'm not advocating for this to replace the standard definition, nor claiming that is equivalent to the standard definition in all possible cases. I am presenting it simply as a novel way to think about the derivative, one that offers deep insight into the forms of the derivatives of certain functions - namely power functions and log functions, both of which are continuously differentiable. First, the algebra is much cleaner for these functions as the x's 'pop out' of the limits. As a result, the power reduction in the power rule is immediately obvious for ANY real non-zero power, and the rule itself can be easily proved for any rational power. In contrast, the standard proof (expanding the binomial) really only works well for positive integers. And for logs, the form of the derivative is seen directly: for any fixed value of t, the 'rise' is constant for all x, and the 'run' is proportional to x, so the derivative of any log function must be proportional to 1/x. We also get a nice closed-form derivation of |x| (which of course is not differentiable at 0) from first principles to boot. These are all concepts that could easily be shown to a bright Calc 1 student or class ALONGSIDE the traditional definition, with the appropriate caveats. But perhaps it's only useful to people who watch videos like this and enjoy revisiting math they've learned to understand it on a deeper level. Maybe I could have framed it differently and spelled out that I was more interested in understanding structure than going for mathematical rigor, as it is a youtube explainer, not a published paper. But that's really all this video is about - changing perspective to gain deeper insight into fundamental concepts.
This was amazing, and you did really address all the important caveats in the video itself. I'm so happy you took the approach of patiently tying it all the way back to the standard definition and showed exactly the right amount of detail and examples to drive home the advantages of using this method in specific cases. I can already see how this can help me in my own research - having another method with a different set of limitations handy can help me discover more properties of the object I'm studying. I'm sure other people will benefit from it too. Thank you!
@@imaginaryangle Thanks for the kind words! A lot of commenters seem hyper focused on certain details and seem to miss the big picture - learning 'why' certain things work the way they do. I'm tempted to make a short addendum video reframing and highlighting the big picture. But maybe I'd just be spinning my wheels. What do you think?
@@HyperCubist I would definitely like to see you explore further in this direction on top of what you've already done, but I honestly think there's nothing to fix about the framing, you've stated very clearly what this is and what it's not. It would be much more interesting to see you talk about more techniques that are overlooked because people are scared a student will shoot themselves in the foot with it 🤭 I love your style and I hope you make more videos like this!
@@incription For exponentiating, the standard definition is the best definition. If f(x)=aˣ, f'(x)=(aˣ⁺ʰ-aˣ)/h=aˣ·f'(0). This relation is really quite neat.
As someone who thinks about pedagogy, I have a few thoughts! As you allude to in the video, this "new" definition of the derivative is really the same definition just with a change of variable. Getting this idea across is actually the crux of problem when it comes to teaching I think. At this stage in their academic careers, many students (in my estimation) still conceptualize substitution and change of variable in terms of a mechanism/algorithm rather than as a fundamental structure-preserving transformation, and this can block the clarity in the intuition that the two formulations are really equivalent, which can make the overall definition harder to grasp.
Strange - I imagine it would be the opposite. Showing that there are multiple algebraic ways of moving one point toward the other to obtain the same result would strengthen the concept itself, and de-emphasize the algebra, and hence the fixed formula, no?
@@HyperCubist I can definitely see it both ways in this case, and the video certainly makes some compelling points, but I'll maintain (maybe just for argument's sake) that learning a new concept is usually a sensitive time where you'd like to present things in as much of a controlled and incremental manner as you can. While an alternative definition here does make the mechanics of solving certain problems much nicer (as you demonstrated in the video) I think more "brute-force" solutions tend to be preferred just because they require less effort to conceptualize. Then again, I'm sure someone else would have a different opinion! At the very least, I see the subject here as excellent supplementary material for a student who has already taken a calculus class (or perhaps is toward the end of one) and is in a good position to re-evaluate and strengthen established concepts.
@@t3ssel8r Yeah both approaches have advantages. You have to learn the brute-force methods certainly, and of course you have to learn to use the original definition. But one thing I often see with my students (I'm a tutor with decades of experience, not a professor) is that even students with solid brute-force algebraic skills often have weak conceptual understanding. They're used to grinding out problems without being able to explain what's happening under the hood or why it works. A lot of what I do is provide that conceptual understanding so that they're able to think about problems at a higher level - things are a lot easier to understand when you know why it works. I think a great way to introduce this would be to practice the original definition first, and once students are adept, show the new definition, and use it to make derivatives of power functions (and the power rule itself) concrete.
My high school calculus taught us using that secant-> tangent method. Is seriously attribute all of my ability to intuitively understand calculus to that man.
I think the most interesting thing to take from this video is that we can use suitable substitutions on the limit definition to get "new definitions" that can be easier for certain problems. It's a useful thing to have on the toolbox, specially for students just starting out. I know I didn't appreciate substitutions enough when entering uni.
Not, really, that is pretty much 90% of mathematics. Just different forms of the same thing. I suggest you start learning category theory now as it deals with the abstract nature of things and how, basically, everything's pretty much the same things in different forms. E.g., think of humans, we are all the same but look different. You don't appreciate anything until you have spend 10-20+ years on it. Math is not what you learn in school, it's what you teach yourself from the real experts. What you learn in school is like learning vocab and grammar... what you learn in life is how to speak and communicate. School does not make you fluent in math(it can if you happen to be really driven and in the perfect environment but all school does is force you to spend time on the subject matter and to push yourself).
@MDNQ-ud1ty while I agree with you, I was thinking more in terms of students not of math directly but of other domains that rely on math, such as engineering (my case). I absolutely love math for the sake of itself, but that's not the case for most of my colleagues. Most engineering students will not tackle category theory and getting out of calc 1 with a good appreciation for substitutions is a game changer. Later, when studying electromagnetism, lots of problems could be solved by having a clever physical consideration that led to a suitable substitution. Having this fluency with subs helps to make these types of connections without higher level maths.
@@victorscarpes Most people are m0r0ns. What you are saying is true in general. Most people will not exert themselves beyond what is necessary to pay the bills(which is also why most rich people are the biggest m0r0ns). Without math there is no engineering or science or modern humanity as we know it. the issue isn't math but the education system which intentionally dumbs people down because the goal for capitalism is to maximize profits, not intelligence or personal awareness. The goal for capitalism is to only teach a person as much as necessary to do the job because why put in a lot of resources to teach them if they are just going to die one day? Substitutions are the foundation of all intelligence so why would it not be a priority in education? without being able reduce complexity(which is what substitutions let one do) humans could do what they do. Most "students" quickly learn that the real world is vastly different than what they were taught. There is a reason why most children have a vast interest in learning things but most adults don't. At some point along the way their desire to be intelligent beings is eliminated. You cannot cow tow to such mentalities... else you become part of that mentality.
Speaking as a professional Calculus tutor, I think that, while some of the calculations are more intuitive, the definition itself is less so. I think the definition is in fact more important. After learning the definition, the student learns the basic functions and the rules for combining them. I think that demonstrating the chain rule through such a definition would be a real brain buster for students.
Even in high school I never had any problems understanding the limit definition. I think it is THE most intuitive way to introduce differentiation to anyone. Yes algebraically it may become messy, but conceptually it is really easy. You measure the slope of a function at a point around an increasingly small interval. And you define the derivative as the value it approaches. The limit h->0 definition combined with an example of a cars speed getting measured with increasingly enhanced accuracy is the kind of 1-2 punch that gets the concept of derivatives across most efficiently.
@@dekippiesipit is a very simple definition and super easy to understand. I don’t understand why people are so dumb. I aced every college course without having to study. Also, I prefer the non-standard infinitesimal approach, not because it is conceptually easier to understand, but because infinitesimals are aesthetically superior to limits
My teacher used to tell us that the right definition for derivation is f(x+y) = f(x) +df(x)(y) ||x-y|| + o(||x-y||) That definition using a notation similar to taylor series always made the most sense to me as it outlines what a derivative really is, it is just a linear approximation of a function. Hence that makes the binomial solution make a lot of sense because you only want to keep first order terms
You can actually extend this idea and use a custom way of approaching x based on the function you need to differentiate. Maybe using (sqrt(x)+h)^2 or e^(ln(x)+h) or some other expression can magically simplify the numerator without being too bad in the denominator
17:00 is amazing. Thank you sooo much. It has been so many years. As a game developer, derivatives matter every now and then in my daily work. 20:00 blew my mind.
Its the same with hyperreal numbers. If you use them in the standard derivative method (as "h"), you do not even need any limit to get your result. All the "dx" magic (multiplying, dividing, etc by "dx") is often times viewed as bad practise. With hyperreal numbers, you can do "anything" with "dx" without invalidating any equation.
I love your video lesson about the derivative. Very neat, clean , very didactic and elegant explanations. I will keep both ways to find the derivative of a function nevertheless interesting to mention that in certain optimization problems We let the computer do the work [for example optimizing an objective function, performance function, cost function or solving and optimal control optimization...]. I will keep visiting your channel for more exciting mathematical definitions, strategies, etc...
I think the warning about limited options should have been put at the beginning, not the end. The "advertisement" at the beginning creates the impression of "super power", but at the end with the "fine print" we realize that the "product" is not perfect. This is math, science - no need marketing and "9.99" magic required. Great video! Great broadening of horizons! Good pedagogical approach and illustrations.
A) I really like this idea for teaching the derivative. I am going to school for mathematics education and I truly believe that this would be more understandable for most students. B) You have a very calming voice! I might use this video to help me sleep in the future 😅
... Good day to you, At time 13.22 Instead of applying u = sqrt(t), we could also consider the denominator (t - 1) as a difference of 2 squares: t - 1 = (sqrt(t) - 1)(sqrt(t) + 1) ... and then cancelling the common factor (sqrt(t) - 1) of top and bottom, resulting in the expression of 1/(sqrt(t) + 1), finally lim(t --> 1)(1/(sqrt(t) + 1) = 1/2; resulting in the same outcome of 1/2sqrt(x). Very nice presentation and thank you for your educational math efforts ... Jan-W
Very nice! Also explaining the limitations of your findings make them even more useful, I’m linking them to a friend of mine who teaches physics in high school, hope it’s ok
I like it. Caveats aside; Its another useful tool, and it has one valuable lesson; sometimes drifting ever so slightly from the railroad tracks we tend to be taught down can open possibilities of shortcuts in certain situations or a better understanding by just looking at the same thing from a different angle. Well done.
1- you will have problems in the examples with the new definition at x=0. 2- at x eq0 you will see that the new and old definition are the same for sufficiently good functions by applying L’Hopital to the new definition
For f(x)=aˣ, the x+h definition gives f'(x)=f'(0)aˣ, which I always found really neat! Never could I have expected similar elegant results for f(x)=xⁿ, f(x)=logₐ(x), and f(x)=|x|. As an IMO gold medal winner (2022) who regularly scrolls through UA-cam, I have seen quite a lot of math, and I don't think of myself as easily impressed. But sometimes, I see an argument like this, and I am completely blown away by its elegance and simplicity.
My favorite comment so far. You get it: the elegance is the thing. Thanks for the kind words. And stay tuned, I think you'll really dig the next video I have planned.
As for the closed form of the deriviative of |x|, I found that by chance when I was first learning calculus. Basically, I noticed that logs often have an absolute value around their input so they can be defined for negative values, but when you take the derivative you find it's still the same because the derivative of ln(x) is equal to the derivative of ln(-x) which is 1/x. That means means the derivative of ln|x| is also 1/x which you can write dln|x| = dx / x. But using the chain rule you can also find dln|x| = d|x| / |x|. Which means we have d|x| / |x| = dx / x or d|x| / dx = |x| / x.
Cool! You can also get it from defining |x| as sqrt(x^2) and using chain rule. It simplifies to x /|x|, which is directly equivalent to |x| / x. I realized that after making the video :)
|x| / x is also the same as signum(x), and if you flip it upside down to get x / |x|, then it's still d|x|/dx = signum(x) for ℝeal numbers x, but it's now _also_ the normalize function for _vectors._ Defining |x| as sqrt(x²) also gives a hint to another generalization: sqrt(x * ~x), which is the magnitude of a complex number or quaternion, and can still be plugged into x / |x| to get x / sqrt(x * ~x) to normalize any complex number, or even several other types of vector.
@@angeldude101 Unfortunately, I don't think |z| is differentiable in complex numbers, as far as I know. For vectors, you can get a form that looks somewhat like this, but you can't get it to look the same because the change in the length only happens parallel to the vector, so you have to do a dot product to get that component. That makes d|x| = dx · x / |x| but you can't get the dx to the other side of the equals sign because there's no inverse for the dot product, so usually you just parameterize by t and find d|x|/dt = (dx/dt) · x / |x|
this is a lovely new definition, arising from a simple shift in point of view: rather than a decreasingly determined "step" [delta], you collapse the "proportional distance" from x to the second point nice!
This is coming from setting h to hx, in the original definition, that is, a relative increment, not an absolute increment, then renaming 1+h=t: (f(x+hx)-f(x))/(hx) = (f(tx)-f(x))/(tx-x). The relative increment is used in Financial Mathematics, where x is the price of a stock, and the derivative is called the (unit-less) delta sensitivity of a financial product (e.g., options) price (represented by function f) that depends on x.
Interesting. Takes me back a few years/decades. Knowing that the aim is to provide an "instantaneous" value I understand the concept of taking a derivative. These days I just use the standard rules but in an idle moment I'll give this approach a try.
An interesting aspect regarding the topic is that you can define the derivative in Banach spaces using the Frechet derivative, which corresponds to a continuos linear functional between the spaces you are working with. In particular, if you have a function that maps from a Hilbert space to R, that function is an element of its dual space. By the Riesz representation theorem, there exists an element in the space that represents the function. If you consider the norm of the Hilbert space as N(x), then for DN(x) h = with < , > the inner product, the representing element is DN(x) = x / N(x). Taking the Hilbert space to be R with the norm N(x) = |x|, it recovers that DN(x) = | - |' (x) = x / |x|.
Quite cool! I'll remember that trick for the derivative of |x|! I would say that a really key insight not discussed here is that of units. Units are not just for physics and engineering, even though it's a pretty inescapable analogy. I'd say that the key here (or the key of the q-derivative) is that you write f'(x) as (unitful stuff)*(unitless stuff). If x were measured in meters, we don't want our complicated math to have units in it, so we don't really want to write "x+h" because that forces h to have units of meters as well! Instead, we can formulate our problem as we vary the unitless parameter t. Then, we write our answer as (unitful stuff)*(unitless function of unitless ratios). I'm bad at phrasing this stuff without using physics language, but it's the same trick you'd use if you're solving the nonlinear ODE L*f''(t)=-g*sin(f(t)) (a pendulum). Even though the problem is nonlinear, it's easy to prove that the period of the pendulum is T=sqrt(L/g)*r(f(0)), where r is a unitless function of the initial angle. If we really really insist that I can't use the word "units" or "meters" (say I'm doing a graduate math course homework problem!), then I can simply rephrase things in terms of rescaling the variables I work with, like choosing to work with t*x instead of h! It also comes in handy when debugging numerical analysis code. But OK, I've ranted enough for one youtube comment, thanks for the stimulating video!!! :)
Yeah the unites aspect does make it somewhat elegant. Btw - new video is up finally. I'd love to hear what you think . Thanks! - HC ua-cam.com/video/aF5mTK4792s/v-deo.html
It's a really nice concept. I do have a comment on the graphics: when writing functions it's better to keep the notation "upwards" instead of italics - it makes it easier to separate variables from function names. In LaTeX specifically (which is what manim uses under the hood), most common functions have macros, e.g. \sin, \cos, \log, \ln, etc. If there isn't, one can always use \text{function}. For exponentials in the natural base, one can write e^{x}, but that again makes an italic e. So instead, I personally define a new command: ewcommand{\eu}{\mathrm{e}}. Then \eu{x} is typeset correctly.
This definition is equivalent to the other (if the limit exists in both case), indeed every way of approching x can be either done by x+h or tx. Since the real topolgy is second countable, in particular every point have a countable open basis, we may choose countable sequence to fully caracterize limit (instead of nets) now you can show every sequence approching x can be done by some t_i that you mutliply with x or by some h_i that you add to x, taking i--> to infinity you get x. Supposing limit of (f(tx)-f(x))/(tx-x) exist then the classic derivative exists and equals the same, you can have problem where the denominator is undefined but you have to change. That's is why I think the first definition of derivative is the best way, in other word the linear approach is better.
It would be (t + i) and (t - i). Also, factoring is only for rational numbers, and the complex plane does that fall under this requirement so it’s technically not a “factor.”
about a year ago i had to prove the derivative of √x and it took me hours, so when it showed up in the video, i was overwhelmed with emotions i hadn't felt in a while. university level maths is no joke, and i'm glad i'm just studying computer science with a little bit of math on the side.
Zero is a tricky problem where there are infinitely many `1/0` values, but only the single zero covering all the subtly different uses of 'zero'. Given maths typically starts with the idea of 'next' (or increment), then yes, the 'h' method is nearest to that. The 't' method works real well for regular folks who haven't yet reach the required level of pedanticism ;-)
This was very enjoyable. Disclaimer: I am neither a mathematician nor a teacher, so my opinion comes with no authority whatsoever. Convergence is hard to teach/learn correctly, especially if a student will be exposed to higher mathematics further in their education. As convergence is defined in terms of distance rather than a ratio, I feel that the traditional approach might be better at building consistent intuition. I'd probably still throw your method in here and there as an illustration of how things can be approached in different ways. It is always nice to see that alternative approaches give consistent results. Also, well done on the manim scripting, the video flowed very smoothly.
Thanks very much! I think I see your point, though if we treat x as a fixed value, we're still doing the exact same thing, reducing the distance between the points to zero. You could directly substitute h = (t-1)x to change between definitions. Then the (fixed) value of x just changes the 'rate' at which h goes to zero. But yes, it's more of a way of demonstrating that any way of bringing the points together accomplishes the result of the derivative, even something like x^k and letting k--> 1. Actually I didn't use Manim, though I'm planning to learn to learn at some point. Just Apple Keynote for the slides/algebra and GeoGebra for the graphs.
@@HyperCubist "But yes, it's more of a way of demonstrating that any way of bringing the points together accomplishes the result of the derivative ..." - This is a really valuable point to learn early on, and I don't think it is taught explicitly in most places. Learning Manim can be a bit of a pain if you are not familiar with programming already, but it is worth the struggle if your current toolset can't do what you need for future videos. I've obviously subscribed so I don't miss anything :)
@@michaeldamolsen Thanks for the sub. Not posting in a while has killed the momentum, not sure if even subscribers are seeing the new video. Pls check it out if you haven't, I'd love to hear what you think! Thanks! - HC ua-cam.com/video/aF5mTK4792s/v-deo.html
Differentiation made me sense to me when I realised it was a higher order function. For f, it gives you d(f, x) = my+c, that is the line/plane/surface that touches f at x when y=x. If you always set y=x, and you inline d, you recover high-school calc.
sin(tx)-sin(x) =2sin((t-1)x/2)cos((t+1)x/2). We can factor a cos(x) out of the limit but this still requires the limit of sin(θ)/θ. Similarly, for the exponential, we can factor e^x and we’re left with a numerator of e^((t-1)x)-1 which still basically requires the same method as the usual proof.
Yes - iirc I worked out that one the same way, but left it out as it wasn't as elegant as the others, and doesn't really offer much of an advantage conceptually or algebraically. Still cool that it works.
Yeah I had come across this quite early on actually, I don't have a formal education in calculus but from the stuff I do know there have been multiple definitions of the derivative I have come across, including the symmetric derivative which is lim h->0 (f(x+h)-f(x-h))/(2h), I found that particular one interesting because it actually gives a value to some discontinuous functions at their discontinuity that make a lot of sense.
Yes! That was exactly the form that got me playing around with the definition and finding the tx form on my own, before learning it was already a thing :)
@@HyperCubist If I remember correctly when I was playing around with the symmetric derivative I also tried to come up with a symmetric version of the multiplicative version and don't remember if I succeeded or not but I do remember that one of the forms I tried was lim h->0 (f(hx)-f(-hx))/(2hx) and also found some variations of this for plugging in higher order polynomials that were interesting in properties.
@@HyperCubist That reminds me that I actually went from what I said there to lim h->0 (f(x(1+h))-f(x(1-h)))/(2hx) and also started plugging in higher orders of h into that to see what happened.
9:40 Regarding a proof of the identity x^n-1=(x-1)(x^(n-1)+...+1), you can also just note that (x^(n-1)+...+1) is a geometric sum and equals (x^n-1)/(x-1) which gives you the result immediately.
As an Engineer who has been exposed to a lot of math in my life, I find this video fascinating and I love this new approach. That said, I don't think that this is how I would have liked to learn derivatives the first time I came across the concept. Delta(y)/Delta(x) which converts to (y1-y)/(x1-x) and from there to [f(x1)-f(x)]/(x1-x) defines the concept. And from there to call x1 = x+delta(x) there is only one step. But it would not even be my immediate step, I would stick with the "moving" point approaching the reference point for a while before moving to the delta approaching zero. For example, for y=x^2, dy/dx = lim |x1->x| (x1^2-x^2)/(x1-x), working with the difference of squares we get lim |x1->x| (x1+x)(x1-x)/(x1-x) = lim |x1->x| (x1+x) and now we can replace x1 with x giving (x+x) = 2x.
That's basically how I'm doing it, except I let x1 = tx, and take the limit as t--> 1. That's what allows the x's to pop outside of the limit, giving you the form right away (and the algebra is cleaner)
@@HyperCubist ... Again, your new approach is great, I love it. The algebra is cleaner indeed at least in many cases. Just that, in my opinion, the addition of the t tending to 1 (or of the delta x tending to 0 for the matter) is a "distraction" in the middle of trying to understand the concept. That's why what is called the "alternative definition" (with a "free" value of x approaching the reference value of x) is the one I like most just at the very beginning. And that is just my subjective opinion.
@@adb012 OK I get what you're saying now. Yes, there's certainly value in that, no question. It's always great to see the same concept done multiple ways. As I recall I first learned lim x-->a f(x)-f(a) / x-a, which is basically the "alternative definition" at a specific point (a). Then we went straight to the standard form. I didn't run into the more general "alternative form" until later. I've tutored calc countless times, and I always have to figure out which way the students class introduces it so I don't confuse them.
Love the video. I enjoyed it very much. Just one minor nuisance. While watching the video for the first time in a stop-motion fashion (i.e., stopping after every point and studying the formulas) I got confused at 20:13 where you convert 1/u * ln(1+u) to ln(1+u)^(1/u) with an animation that seems to indicate [ln(1+u)]^(1/u) since the first expression is log of (1+u) and the animation just moves the 1/u part right above the entire expression with out any other hints. After a while I figured out that this is not what you intended but it would have been easier to follow if you could add some extra brackets, as in ln[(1+u)^(1/u)] or some other indicator. Again, thanks for a good video.
Maybe the derivative isn't usually defined in this way because this doesn't generalize to higher dimensions all that well? We usually require the convergence along all paths, not just "linear" ones (tx only hits multiples of x)
Btw, we can substitute h=(t-1)x = tx-x as the x is constant for each limit. If t approaches 1, h will approach 0 and we get our usual definition of the derivative. So we have "If the new derivative exists, then so does the usual". For real x not equal to 0, the converse is also true with t=(h+x)/x. For vectors x, this substitution doesn't work anymore tho (as we cant divide by vectors).
@@reeeeeplease1178 "For vectors x, this substitution doesn't work anymore tho (as we cant divide by vectors)." Only if you don't try hard enough. ;) 1/v⃗ = v⃗¯¹ = (~v⃗)/|v⃗|² = (~v)/(v⃗ * ~v⃗) (~v⃗ is kind of like a complex conjugate, though for several inputs is also an identity function.) If you're curious how this way of multiplying and dividing by vectors works out, look up Clifford/Geometric Algebra.
Thank you! I finally understand how calculating a derivative works! It's real nice how you show multiple examples with both the conventionally standard method and your alternative! Watching this is like practising along almost.
I came up with similar unorthodox methods of computing derivatives using the limit definition and was I was rather scoffed off by my peers and shunned by the teachers, I expected them to be a bit more supportive and guide me thru the technicalities and making the process more rigourous. But oh well we dont have time for that when we're preparing for competitive exams🙅♀️
@@HyperCubist I had already watched it this week earlier ✨ and It did refresh my intuition of the exponential function, I love your approach to the typical "math concept explainer videos", the steps you take feel as the obvious next step and make it seem super natural. The cohesiveness and coherence of the videos is great. Awesome content 💫💫 would love to see more from you!
Very cool! As someone that gets the standard form conceptually, while also being rubbed entirely the wrong way by it, having this other option makes me feel much less anxiety at the prospect of doing them but hand... Would be cool to see a similar alternative for integration or some other related topics
Pretty nice video! Minor reflection. It seems like for the examples you essentially get to replace binomial expansion and factoring an h with doing polynomial division. You kind of hand waved the polynomial division. But that doesn’t actually seem simpler. It’s not exactly harder either. But I see little gain here. Also, this totally breaks when x=0. Finally, you might argue that the more difficult bit about derivatives is the concept of a limit and rules regarding it. That’s because it’s usually not very clearly defined at this level. The sqrt, abs and log results were pretty nice. Always nice to see multiple perspectives.
I mentioned the zero part at the end, but it does break down for certain types of functions (see pinned comment). The polynomial factoring can be learned by a simple pattern, or you can prove it with synthetic division, or just verify it by multiplying. What I should have mentioned is that calc students already know that factoring, as it's just a rearrangement of the sum of a finite geometric series. Compare a one variable factoring formula where you know every term to the (x+h) binomial form, with two variables, and where most terms are difficult to write for a general power n.
I can see this working like a math experiment in a classroom-setting. First, you demonstrate that the secant approaches the tangent (I think Newton’s lemmas in his Principia work well for this step). And then you ask how that might play itself out mathematically, and let your students play around with different models. This multiplication definition might arise, even though I would the majority of solvers to figure out the addition version first.
That would be an interesting experiment indeed. Btw - new video is up finally. I'd love to hear what you think . Thanks! - HC ua-cam.com/video/aF5mTK4792s/v-deo.html
Essentially if ii understood well, what just happened is that the 't' factor defines the form of the coefficient to the X^n-1 term while the x term(exponent carrier) defines the form of the main derivative. we could even render that more rigourous by studying all such forms gotten from the algebraic mouvements expressed by factor 't' if we define some set "C" for (coefficient ) that can be adjoint to some set "F" for (Forms); all such forms carried by the main exponent carrier x. This could help us meausure more rigourously advantages in speed and other factors such as cleaniness of steps that arise due to the two methods. Donot know if all what i said makes sense but if not, we here to talk and contribute. Those were my thoughts.
Yooo, this is super dope! I've never really thought about redefining the limit-definition of the derivative to make it lend itself better algebraically to certain functions. Absolutely eye-opening and interesting! Now I'm wondering about perhaps other ways of extending the definition... (E.g. instead of an additive or a multiplicative change in `x`, perhaps one considers a different change in x, such that the algebra works out. In general, you'd have `z = g(x, h)`, where `g(x,0)=x`, and `g` takes whatever form is convenient for what you're trying to differentiate).
Thanks so much! Right - there are lots of ways you can define it. I was messing with z = x^k, and letting k --> 1, but couldn't find any functions that benefited from it. Also you can let both points move towards a single point. A variation that sometimes shows up in exercises is using x-h and x+h, with 2h in the denominator. If you find anything cool let me know!
@@HyperCubist The x+h & x-h method can be separated out as the average of the right-hand & left-hand derivatives. If the limit exists, the left & right limits must be equal, so the average *is* that limit. As shown when you did the u-substitution, any method that makes the 2 points approach in the limit should work. In fact, all these methods can be summarized as f'(x)=[f(g(x)-f(x)]/(g(x)-x), for some g(x) approaching x.
Unfortunately not posting in a while killed the juice. Would love it if you checked out the new vid - thanks! - HC ua-cam.com/video/aF5mTK4792s/v-deo.html
Unfortunately not posting in a while has displeased the algorithm. But thenNew video is up finally. I'd love to hear what you think . Thanks! - HC ua-cam.com/video/aF5mTK4792s/v-deo.html
The reason it's not taught is really simple. Mathematician love generality, but the new definition only works for Real functions. In multivariable calculus, all paths have to have the same limit point, but with f(tx)-f(x) there is only one path possible: along the direction x points to (the existence of partial or directional derivatives does not guarantee a total derivative). In multi.dim. it has to be possible to independently choose a point and a direction and now it's the typical derivative or a directional derivative in the 1D case, since the only positive oriented normed basis vector is 1 itself.
You're right, it doesn't generalize to multivariable. The point is that it can be used as a pedagogical tool for learning power rule on a deeper level, and simplifies beautifully for several types of functions.
I was taught a different way to calculate differentiate functions using h terms and you take O(.) of the terms to find the derivative. I don't know what this is called but I can find a reference if anyone is interested.
Very well-put together! Personally I'm not entirely convinced of its use---to me, a lot of the nontrivial examples end up needing tricks or arguments that are roughly equivalent in trickiness/difficulty as in the original method's proofs anyway :)) So I don't really see any advantages---through intuition or ease of rigor---that this alternative definition provides. But I can still appreciate a video with clear explanations and great visuals, and this video is certainly excellent! The video was paced quite well and you have a voice well-suited for voiceovers. Some particular comments I have: -Re: |x|. Gonna start with a pet peeve of mine---people not liking piecewise functions and somehow thinking "closed forms" are superior. I would prefer to disabuse people of this notion as soon as possible, actually!!! Is there any particular reason for |x|/x to be preferable over the +/-1 piecewise? I'd argue no, and In fact, I'd argue it's worse, because it's just needlessly obfuscating the simple actual intuitive thing going on (-1 slope when negative, +1 slope when positive). Justice for if statements!! Elementary closed forms are tooooootally overrated, and I don't want to encourage my students to put them on even more of a pedestal :)) Also, computing the derivative of |x| using the standard definition is not at all hard. The only technical bit is needing to argue that when x!=0, we know that |x+t| and |x| have the same sign when t is sufficiently small----an argument that your alternate solution also still basically had to make, by the way, when you argued that |t| = t as t -> 1 :PP -Re: sqrt(x). Again, your argument is of equal complexity to just using the standard definition of the derivative anyway. You still had to use the "conjugate trick" in the end to ultimately resolve your limit expression (while the substitution trick is clever, that's... just doing the conjugate trick, but wearing a fake mustache). - Re: Power rule. I really don't see what's so unintuitive with the "binomial expansion". In fact, the argument here is the beating heart of understanding calculus intuitively: recognizing that (x+h)^n = x^n + nx^(n-1) h + (a bunch of other terms that aren't important because they go to 0). In fact, it feels disingenuously "over-spooky" to say that it _needs_ the binomial expansion and to write out all the binomial coefficients (you even expanded C(n, 2) into n(n-1)/2, what is the point of that other than to scare people??), when most of the coefficients are actually completely irrelevant and don't need to be written out. (What follows is a somewhat weird take I have which is off-topic from general Calc 1 pedagogy) Also: you can prove the power rule for rational exponents i.e. x^(a/b) by proving x^a and x^(1/b) separately and then invoking the chain rule. And then it's not immediately obvious how to extend this to real numbers... which I like! Because it forces students to think critically about what the heck something like 2^pi is _even supposed to mean_!!! I hate how much weirdness about real numbers just gets swept under the rug in math classes, since many students (my past self included) end up with the feeling that real numbers are totally normal things, when actually they are Incredibly Weird and that rigorously defining things with them is Incredibly Tricky And yes, I'm talking about rigor here. Because if we disregard rigor, I'm pretty sure most people would already be convinced that the power rule holds in general, because it holds for integers and rational exponents, and like---polynomials are nice, so i wouldn't expect them to be too weird and suddenly break. So the only time (imho) we're ever actually discussing details is when in search of rigorous definitions and proofs. My understanding of the typical proof of power rule for real-number exponents is to let y = x^t imply ln(y) = t * ln(x) and then do implicit differentiation---which isn't so tough at all! Which finally leads me to... - Re: ln(x). In many calculus classes, ln and exp are held off until the integral lesson, and then it is _defined_ to be the antiderivative of 1/x and e^x is defined as its inverse (from which many familiar properties of it and exp(x) can be derived). Once again, I emphasize that you somewhat _need_ to do this because otherwise there is this gigantic gaping hole of what a real-number exponent _even means at all_. Your proposed solution here takes a shortcut by just taking for granted that ln(x) and exp(x) exist and are things we can use... when they really haven't been defined yet at this point in time, and to be frank "I have no idea what they mean" :P My preferred intuition is to define exp(x) by its taylor series, but obviously that can't be used until Calc 2. I suppose you could go define e.g. 2^pi as the limit of the sequence 2^3, 2^3.1, 2^3.14, 2^3.141, 2^3.1415... but that involves a lot of handwaving about continuity that makes the argument feel weaker and more wishy-washy (freshman me sure wouldn't have been ready to deal with Cauchy sequences back then!!!)
Wow, thanks for the kind words and thoughtful reply! I won't hit every point, but... The big advantage (as I see it) is that (for these examples - power functions, logs, and |x|) all the factors of x 'pop out' right away before having to simplify the limit - so with almost no work you get to see what the derivative looks like to within a constant. Then you only have to solve a limit, with one variable rather than two. I emphasize this more in the follow up video (on my channel), but it's also a question of understanding intuitively WHY power rule reduces power functions by one degree, which can be seen for any real power, not just integers/rationals. So even if a student can't fully prove power rule for x^pi right from the get-go, they can understand exactly why it's proportional to x^(pi-1), using first principles (rise over run) - no chain rule or implicit diff to act as a barrier to understanding. The same logic applies to ln(x) - you can see exactly why rise/run is proportional to 1/x, and watch as it pops out of the limit. As for if it's actually easier, try this. Use the standard definition ONLY (first principles - no chain rule) to find the derivative of x^5/7. I tried it myself the other day, and you CAN do it, but like the conjugate, you have to pull a factor expression out of thin air, and it's a god awful mess, nothing I would expect a student to do. Now try it with the 'tx' form. One substitution later and you're factoring simple polynomials (u^n - 1) that follow a simple and easily learnable form - one that students have technically already learned with finite geometric series. Then try the general rational case: x^m/n, or even x^-m/n, using both definitions. Re: the conjugate trick - standard and useful as it may be - requires supplying factors from nothing, as in the example above. The substitution trick doesn't, you just have to factor what's already in front of you. As for the pedagogical stuff and real powers, all I know is that students have been working with e^x and ln(x) since pre-calc, and treating them as continuous functions, so they must have some level of comfort with real powers.
Some comments: * When deriving the derivative of |x|, I think you did use the piecewise definition when you said as t->1, |t| is t. * For the ln(x) derivative, you went from needing the chain rule to needing to know when to interchange logs and limits as well as knowing how to expand the series (1+1/n)^n.
@@HyperCubist although the new definition you found does make it a lot easier to work out. I just figured out how to work it out using the limit definition while I was evaluating some functions that made use of the absolute value.
Wonderful video! Incredible cross over of maths and satire. I haven't laughed so hard in ages. The slow burn of each example had me rolling! Keep it up!
This just goes to show that school was never meant to teach you. If this definition was introduced to my calculus class, so many more students would have understood
I like your new definition. Well, I like all math. I was exited to see how the h->1 defination would work for e^x. In the end, still a nice video. Good graphics and examples.
I suppose this would be a good way to re-introduce limits in multivariable calc when considering limits along different paths. As far as intro Calc, doubtful it will really make things any simpler or easier to grasp.
I've derived a similar closed form version of the derivative of absolute value of x by using abs(x) = sqrt(x^2) which is the vector way of writing it lol. Interestingly, you get the fraction the other way around, as x/|x| instead.
Right - I came across that when playing around with it as well. It does use chain rule though - this way uses the actual definition of the derivative directly.
As a challenge problem - try using this definition to prove the power rule for rational exponents: f(x) = x^(m/n). It's something the standard definition doesn't do easily.
@@HyperCubist Got it, tried a u = t^m/n substitution first but that didn't work but then I did u = t^1/n and that gave both the numerator and denominator integer powers - thanks for the tip!
Have you seen John Gabriel's 'New Calculus'? I like his as well but it is radically different as it doesn't use limits at all. The definition is purely algebraic as it relies on the exact definition of the tangent line. It essentially combines left and right handed derivation into one.
Very nice video, I would have really enjoyed a proof that the two are equivalent, I think it is enough to say that it has the product rule and the derivative of x is 1 (because then it is easy to have the derivative on all polynomials, including Taylo approximations of C^2 functions).
It was already 23 minutes long lol! But they are equivalent except at x = 0, in which case there's the lim x-->0 path. And even then there are some weird exceptions (see the note in the link). But you can prove all the standard rules in similar ways as the x+h def. Keep in mind it's meant as an elegant tweak, not a replacement.
I was hoping for you to mention that this method is like saying f'(x) = g(x).f'(1) Where g(x) is generally easier to find in shown cases, and f'(1) is some limit we'd have to solve 😅
You know someone else made that same comment, and I was scratching my head trying to figure out why that's interesting? I mean you get a function of x, times a limit, which is constant. What's the significance that it happens to be f'(1)? I could see that being interesting for f(x) = a^x using the old definition, as f'(x) = f(x)*f'(0); it's proportional to the original function. But in this case the derivative is proportional to a different function, so what's the significance? Help me out - I'm legit curious.
A reason why I think this definition is used less is that it doesn't generalize well to multiple dimensions. As soon as you have multiple dimensions your limit needs to be able to approach for every direction. In 1d this is fine since q can fluctuate as q1. In multiple dimensions you would need to define some multiplication on vectors which can be quite tricky.
It's completely possible to define a product on vectors. The best version that I've seen has been to define a bilinear associative product such that a vector squared is a scalar. However this does mean the result is a noticeably different kind of object than you started with, and also that it's non-commutative in general, so it would be interesting to see how this version would work with such a product. Then again, wouldn't the traditional definition also need to divide by a vector if h is meant to approach 0 from all directions?
I think if you use sum (difference) to product formulas, you can get the derivatives of sine and cosine. But you would still need to know sin(u)/u -> 1 as u ->0.
If you're willing to use implicit differentiation and the arc length formula, you can derive the derivatives of sine and cosine in terms of each other without proving that sin(u)/u goes to 1 as u goes to 0.
@@doruk1713 I don't know if you were asking me as I think @matthuelson3409 had the more interesting comment. But if you were, here it is (without carrying the limits): sin(A) - sin(B) = 2sin([A-B]/2)cos([A+B]/2) so [sin(tx) - sin(x)]/(tx-x) = 2sin([tx-x]/2)cos([tx+x]/2)/(tx-x) = cos([tx+x]/2)*sin([tx-x]/2)/[(tx-x)/2] -> cos(2x/2)* sin(u)/u -> cos(x) cos(A) - cos(B) = -2sin([A+B]/2)sin([A-B]/2) so [cos(tx) - cos(x)]/(tx-x) = -2sin([tx+x]/2)sin([tx-x]/2)/(tx-x) = -sin([tx+x]/2)*sin([tx-x]/2)/[(tx-x)/2] -> -sin(2x/2)* sin(u)/u -> -sin(x)
The quick and dirty way: Use x^2 + y^2 = 1 and (x')^2 + (y')^2 =1 (this is parametrization by arc length). Also, we use initial conditions x(0)=1 and y'(0)>0, so the curve is traced counterclockwise. Take derivatives of x^2 + y^2 = 1, implicitly: 2x x'+2y y' = 0, implying y y' = -x x' and squaring this yields y^2 y'^2 =x^2 x'^2. Now, multiply both sides of (x')^2 + (y')^2 =1 by x^2, obtaining x^2 x'^2 +x^2 y'^2 = x^2. Replace x^2 x'^2 with y^2 y'^2, obtaining y^2 y'^2 +x^2 y'^2 = x^2. Factor out a y'^2 on the left: (y^2 + x^2) y'^2 = x^2, and, since y^2 +x^2 =1, we arrive at y'^2 = x^2. The initial conditions then yield y' = x. Interpret y = sin t and x = cos t (we are on the unit circle, after all), and we're home free. There are subtleties to be ironed out, mostly around continuity and taking square roots properly.
Yes, it doesn't extend in the general case. Just another tool in the toolbox for certain 1D functions, great for building up intuition. Btw - New video is up finally. I'd love to hear what you think . Thanks! - HC ua-cam.com/video/aF5mTK4792s/v-deo.html
I thought about something like one day, but never push it, and I'm glad you did because it's infact interresting ! To me, both definitions are good, it's just a matter of "does the function I want to know the derivative of behave more nicely with addition or multiplication ?" ; of course power functions behave much better with multiplication since for them f(x*y) = f(x)*f(y), and exponentials behave much better with addition since f(x + y) = f(x)*f(y). I'm wondering now if one could take even another definition for "z", adapting this definition to be the best suited for the function considered ; for example if you want to know the derivative of a function "f" for which f(x^y) = f(y)/f(y) , could we create the definition of derivative better suited for the calculation ?... If any ideas please share ^^ And really cool video :)
Just leaving a comment to address some common objections. Yes, this is the limiting case of the quantum derivative (mentioned at the end). No, it does not extend well to higher dimensions. Yes, the multiplicative definition is undefined at zero. There is a fix for this mentioned at the end, f'(0) = lim x--> 0 of f'(x), if this limit exists. But an an important caveat that I did overlook, is that even this 'patch' only works for functions that are continuously differentiable on their domain - an exception would be something like f(x) = x^2 sin(1/x).
More importantly, I'm not advocating for this to replace the standard definition, nor claiming that is equivalent to the standard definition in all possible cases. I am presenting it simply as a novel way to think about the derivative, one that offers deep insight into the forms of the derivatives of certain functions - namely power functions and log functions, both of which are continuously differentiable.
First, the algebra is much cleaner for these functions as the x's 'pop out' of the limits. As a result, the power reduction in the power rule is immediately obvious for ANY real non-zero power, and the rule itself can be easily proved for any rational power. In contrast, the standard proof (expanding the binomial) really only works well for positive integers. And for logs, the form of the derivative is seen directly: for any fixed value of t, the 'rise' is constant for all x, and the 'run' is proportional to x, so the derivative of any log function must be proportional to 1/x. We also get a nice closed-form derivation of |x| (which of course is not differentiable at 0) from first principles to boot.
These are all concepts that could easily be shown to a bright Calc 1 student or class ALONGSIDE the traditional definition, with the appropriate caveats. But perhaps it's only useful to people who watch videos like this and enjoy revisiting math they've learned to understand it on a deeper level. Maybe I could have framed it differently and spelled out that I was more interested in understanding structure than going for mathematical rigor, as it is a youtube explainer, not a published paper. But that's really all this video is about - changing perspective to gain deeper insight into fundamental concepts.
This was amazing, and you did really address all the important caveats in the video itself. I'm so happy you took the approach of patiently tying it all the way back to the standard definition and showed exactly the right amount of detail and examples to drive home the advantages of using this method in specific cases. I can already see how this can help me in my own research - having another method with a different set of limitations handy can help me discover more properties of the object I'm studying. I'm sure other people will benefit from it too. Thank you!
@@imaginaryangle Thanks for the kind words! A lot of commenters seem hyper focused on certain details and seem to miss the big picture - learning 'why' certain things work the way they do. I'm tempted to make a short addendum video reframing and highlighting the big picture. But maybe I'd just be spinning my wheels. What do you think?
@@HyperCubist I would definitely like to see you explore further in this direction on top of what you've already done, but I honestly think there's nothing to fix about the framing, you've stated very clearly what this is and what it's not. It would be much more interesting to see you talk about more techniques that are overlooked because people are scared a student will shoot themselves in the foot with it 🤭 I love your style and I hope you make more videos like this!
The reason for calling coefficients constants is the kernel of the illusion?
"[...] people who watch videos like this and enjoy revisiting math they've learned to understand it on a deeper level."
Yes. 🙂
as a rule of thumb, if your function plays nicely with addition, use the original, and if it plays nicely with multiplication, use this new one
There must be a family of other methods that work well with other functions
what about exponentiation?
@@Kavukamariyou cats are COOKING 🎷
@@KavukamariI'll give you $100 if you find one that works well with a different family of functions
@@incription For exponentiating, the standard definition is the best definition.
If f(x)=aˣ,
f'(x)=(aˣ⁺ʰ-aˣ)/h=aˣ·f'(0).
This relation is really quite neat.
As someone who thinks about pedagogy, I have a few thoughts! As you allude to in the video, this "new" definition of the derivative is really the same definition just with a change of variable. Getting this idea across is actually the crux of problem when it comes to teaching I think. At this stage in their academic careers, many students (in my estimation) still conceptualize substitution and change of variable in terms of a mechanism/algorithm rather than as a fundamental structure-preserving transformation, and this can block the clarity in the intuition that the two formulations are really equivalent, which can make the overall definition harder to grasp.
Oh hey would you look at that, my favorite game dev youtuber also watched this 😁
Strange - I imagine it would be the opposite. Showing that there are multiple algebraic ways of moving one point toward the other to obtain the same result would strengthen the concept itself, and de-emphasize the algebra, and hence the fixed formula, no?
omg I'm on PedagogyTube, perfect!
@@HyperCubist I can definitely see it both ways in this case, and the video certainly makes some compelling points, but I'll maintain (maybe just for argument's sake) that learning a new concept is usually a sensitive time where you'd like to present things in as much of a controlled and incremental manner as you can. While an alternative definition here does make the mechanics of solving certain problems much nicer (as you demonstrated in the video) I think more "brute-force" solutions tend to be preferred just because they require less effort to conceptualize. Then again, I'm sure someone else would have a different opinion!
At the very least, I see the subject here as excellent supplementary material for a student who has already taken a calculus class (or perhaps is toward the end of one) and is in a good position to re-evaluate and strengthen established concepts.
@@t3ssel8r Yeah both approaches have advantages. You have to learn the brute-force methods certainly, and of course you have to learn to use the original definition. But one thing I often see with my students (I'm a tutor with decades of experience, not a professor) is that even students with solid brute-force algebraic skills often have weak conceptual understanding. They're used to grinding out problems without being able to explain what's happening under the hood or why it works. A lot of what I do is provide that conceptual understanding so that they're able to think about problems at a higher level - things are a lot easier to understand when you know why it works. I think a great way to introduce this would be to practice the original definition first, and once students are adept, show the new definition, and use it to make derivatives of power functions (and the power rule itself) concrete.
My high school calculus taught us using that secant-> tangent method. Is seriously attribute all of my ability to intuitively understand calculus to that man.
I think the most interesting thing to take from this video is that we can use suitable substitutions on the limit definition to get "new definitions" that can be easier for certain problems. It's a useful thing to have on the toolbox, specially for students just starting out. I know I didn't appreciate substitutions enough when entering uni.
Not, really, that is pretty much 90% of mathematics. Just different forms of the same thing.
I suggest you start learning category theory now as it deals with the abstract nature of things and how, basically, everything's pretty much the same things in different forms. E.g., think of humans, we are all the same but look different.
You don't appreciate anything until you have spend 10-20+ years on it. Math is not what you learn in school, it's what you teach yourself from the real experts. What you learn in school is like learning vocab and grammar... what you learn in life is how to speak and communicate. School does not make you fluent in math(it can if you happen to be really driven and in the perfect environment but all school does is force you to spend time on the subject matter and to push yourself).
@MDNQ-ud1ty while I agree with you, I was thinking more in terms of students not of math directly but of other domains that rely on math, such as engineering (my case). I absolutely love math for the sake of itself, but that's not the case for most of my colleagues. Most engineering students will not tackle category theory and getting out of calc 1 with a good appreciation for substitutions is a game changer. Later, when studying electromagnetism, lots of problems could be solved by having a clever physical consideration that led to a suitable substitution. Having this fluency with subs helps to make these types of connections without higher level maths.
@@victorscarpes Most people are m0r0ns. What you are saying is true in general. Most people will not exert themselves beyond what is necessary to pay the bills(which is also why most rich people are the biggest m0r0ns).
Without math there is no engineering or science or modern humanity as we know it. the issue isn't math but the education system which intentionally dumbs people down because the goal for capitalism is to maximize profits, not intelligence or personal awareness. The goal for capitalism is to only teach a person as much as necessary to do the job because why put in a lot of resources to teach them if they are just going to die one day?
Substitutions are the foundation of all intelligence so why would it not be a priority in education? without being able reduce complexity(which is what substitutions let one do) humans could do what they do.
Most "students" quickly learn that the real world is vastly different than what they were taught. There is a reason why most children have a vast interest in learning things but most adults don't. At some point along the way their desire to be intelligent beings is eliminated.
You cannot cow tow to such mentalities... else you become part of that mentality.
@MDNQ-ud1ty School doesn't make you fluent in math and neither did the 10 minute video you watched on category theory. Okay buddy? Relax.
@@grivza You are an 1mb3cile.
I've actually read CFTWM along with CM and many others. But keep telling yourself your 93 IQ is an A.
Speaking as a professional Calculus tutor, I think that, while some of the calculations are more intuitive, the definition itself is less so. I think the definition is in fact more important. After learning the definition, the student learns the basic functions and the rules for combining them. I think that demonstrating the chain rule through such a definition would be a real brain buster for students.
Even in high school I never had any problems understanding the limit definition. I think it is THE most intuitive way to introduce differentiation to anyone. Yes algebraically it may become messy, but conceptually it is really easy.
You measure the slope of a function at a point around an increasingly small interval. And you define the derivative as the value it approaches. The limit h->0 definition combined with an example of a cars speed getting measured with increasingly enhanced accuracy is the kind of 1-2 punch that gets the concept of derivatives across most efficiently.
It is literally the same definition, lmao. Substitution of variables doesn’t change the definition
@@dekippiesipit is a very simple definition and super easy to understand. I don’t understand why people are so dumb. I aced every college course without having to study.
Also, I prefer the non-standard infinitesimal approach, not because it is conceptually easier to understand, but because infinitesimals are aesthetically superior to limits
This is EXACTLY the video I needed to understand the derivative!! I have never understood it before… but now I realize how simple it is!
I'm so glad it helped!
Btw - new video is up finally. I'd love to hear what you think. Thanks! ua-cam.com/video/aF5mTK4792s/v-deo.html
My teacher used to tell us that the right definition for derivation is
f(x+y) = f(x) +df(x)(y) ||x-y|| + o(||x-y||)
That definition using a notation similar to taylor series always made the most sense to me as it outlines what a derivative really is, it is just a linear approximation of a function. Hence that makes the binomial solution make a lot of sense because you only want to keep first order terms
New video is up finally. It's actually related to Taylor series. I'd love to hear what you think. Thanks! ua-cam.com/video/aF5mTK4792s/v-deo.html
You can actually extend this idea and use a custom way of approaching x based on the function you need to differentiate. Maybe using (sqrt(x)+h)^2 or e^(ln(x)+h) or some other expression can magically simplify the numerator without being too bad in the denominator
New video is up finally. I'd love to hear what you think. Thanks! ua-cam.com/video/aF5mTK4792s/v-deo.html
17:00 is amazing. Thank you sooo much. It has been so many years. As a game developer, derivatives matter every now and then in my daily work. 20:00 blew my mind.
Thanks for the kind words. And the new video is up finally. I'd love to hear what you think - HC ua-cam.com/video/aF5mTK4792s/v-deo.html
Its the same with hyperreal numbers. If you use them in the standard derivative method (as "h"), you do not even need any limit to get your result. All the "dx" magic (multiplying, dividing, etc by "dx") is often times viewed as bad practise. With hyperreal numbers, you can do "anything" with "dx" without invalidating any equation.
New video is up finally. I'd love to hear what you think. Thanks! ua-cam.com/video/aF5mTK4792s/v-deo.html
I love your video lesson about the derivative. Very neat, clean , very didactic and elegant explanations. I will keep both ways to find the derivative of a function nevertheless interesting to mention that in certain optimization problems We let the computer do the work [for example optimizing an objective function, performance function, cost function or solving and optimal control optimization...]. I will keep visiting your channel for more exciting mathematical definitions, strategies, etc...
Thanks! I have several calc related videos planned, as well as a while series on 4D stuff.
New video is up finally. I'd love to hear what you think. Thanks! ua-cam.com/video/aF5mTK4792s/v-deo.html
I think the warning about limited options should have been put at the beginning, not the end. The "advertisement" at the beginning creates the impression of "super power", but at the end with the "fine print" we realize that the "product" is not perfect. This is math, science - no need marketing and "9.99" magic required.
Great video! Great broadening of horizons! Good pedagogical approach and illustrations.
Thanks! New video is up finally. I'd love to hear what you think . Thanks! - HC ua-cam.com/video/aF5mTK4792s/v-deo.html
A) I really like this idea for teaching the derivative. I am going to school for mathematics education and I truly believe that this would be more understandable for most students.
B) You have a very calming voice! I might use this video to help me sleep in the future 😅
Thanks so much! And the ew video is up finally. I'd love to hear what you think. ua-cam.com/video/aF5mTK4792s/v-deo.html
... Good day to you, At time 13.22 Instead of applying u = sqrt(t), we could also consider the denominator (t - 1) as a difference of 2 squares: t - 1 = (sqrt(t) - 1)(sqrt(t) + 1) ... and then cancelling the common factor (sqrt(t) - 1) of top and bottom, resulting in the expression of 1/(sqrt(t) + 1), finally lim(t --> 1)(1/(sqrt(t) + 1) = 1/2; resulting in the same outcome of 1/2sqrt(x). Very nice presentation and thank you for your educational math efforts ... Jan-W
Quite true. I did it the way as when you try it for cube roots, nth roots, etc., it's much easier to substitute u = x^1/n. Thanks for the kind words!
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Very nice! Also explaining the limitations of your findings make them even more useful, I’m linking them to a friend of mine who teaches physics in high school, hope it’s ok
I like it. Caveats aside; Its another useful tool, and it has one valuable lesson; sometimes drifting ever so slightly from the railroad tracks we tend to be taught down can open possibilities of shortcuts in certain situations or a better understanding by just looking at the same thing from a different angle. Well done.
Thank you! Great way to put it.
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1- you will have problems in the examples with the new definition at x=0.
2- at x
eq0 you will see that the new and old definition are the same for sufficiently good functions by applying L’Hopital to the new definition
For f(x)=aˣ, the x+h definition gives f'(x)=f'(0)aˣ, which I always found really neat!
Never could I have expected similar elegant results for f(x)=xⁿ, f(x)=logₐ(x), and f(x)=|x|.
As an IMO gold medal winner (2022) who regularly scrolls through UA-cam, I have seen quite a lot of math, and I don't think of myself as easily impressed.
But sometimes, I see an argument like this, and I am completely blown away by its elegance and simplicity.
My favorite comment so far. You get it: the elegance is the thing. Thanks for the kind words. And stay tuned, I think you'll really dig the next video I have planned.
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In quantum calculus, the quantum derivative is actually very helpful in certain circumstances
Underrated comment. That’s really what he is talking about. Probably a physics student.
@@edisondiiorio2642not really, go watch his follow up video
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@@HyperCubistyour feelings would be irrational
As for the closed form of the deriviative of |x|, I found that by chance when I was first learning calculus. Basically, I noticed that logs often have an absolute value around their input so they can be defined for negative values, but when you take the derivative you find it's still the same because the derivative of ln(x) is equal to the derivative of ln(-x) which is 1/x. That means means the derivative of ln|x| is also 1/x which you can write dln|x| = dx / x. But using the chain rule you can also find dln|x| = d|x| / |x|. Which means we have d|x| / |x| = dx / x or d|x| / dx = |x| / x.
Cool! You can also get it from defining |x| as sqrt(x^2) and using chain rule. It simplifies to x /|x|, which is directly equivalent to |x| / x. I realized that after making the video :)
|x| / x is also the same as signum(x), and if you flip it upside down to get x / |x|, then it's still d|x|/dx = signum(x) for ℝeal numbers x, but it's now _also_ the normalize function for _vectors._
Defining |x| as sqrt(x²) also gives a hint to another generalization: sqrt(x * ~x), which is the magnitude of a complex number or quaternion, and can still be plugged into x / |x| to get x / sqrt(x * ~x) to normalize any complex number, or even several other types of vector.
@@angeldude101
Unfortunately, I don't think |z| is differentiable in complex numbers, as far as I know.
For vectors, you can get a form that looks somewhat like this, but you can't get it to look the same because the change in the length only happens parallel to the vector, so you have to do a dot product to get that component.
That makes d|x| = dx · x / |x| but you can't get the dx to the other side of the equals sign because there's no inverse for the dot product, so usually you just parameterize by t and find d|x|/dt = (dx/dt) · x / |x|
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That derivation of the derivative of the natural log was remarkable. As a mere student, congratulations!!
This is the derivation using the usual definition but with one extra step.
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this is a lovely new definition, arising from a simple shift in point of view: rather than a decreasingly determined "step" [delta], you collapse the "proportional distance" from x to the second point
nice!
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This is coming from setting h to hx, in the original definition, that is, a relative increment, not an absolute increment, then renaming 1+h=t: (f(x+hx)-f(x))/(hx) = (f(tx)-f(x))/(tx-x). The relative increment is used in Financial Mathematics, where x is the price of a stock, and the derivative is called the (unit-less) delta sensitivity of a financial product (e.g., options) price (represented by function f) that depends on x.
Awesome video! I love the idea of teaching this alongside the classic definition.
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I have never seen this form of the derivative. Thank you for sharing.
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Interesting. Takes me back a few years/decades. Knowing that the aim is to provide an "instantaneous" value I understand the concept of taking a derivative. These days I just use the standard rules but in an idle moment I'll give this approach a try.
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An interesting aspect regarding the topic is that you can define the derivative in Banach spaces using the Frechet derivative, which corresponds to a continuos linear functional between the spaces you are working with. In particular, if you have a function that maps from a Hilbert space to R, that function is an element of its dual space. By the Riesz representation theorem, there exists an element in the space that represents the function. If you consider the norm of the Hilbert space as N(x), then for DN(x) h = with < , > the inner product, the representing element is DN(x) = x / N(x). Taking the Hilbert space to be R with the norm N(x) = |x|, it recovers that DN(x) = | - |' (x) = x / |x|.
Quite cool! I'll remember that trick for the derivative of |x|! I would say that a really key insight not discussed here is that of units. Units are not just for physics and engineering, even though it's a pretty inescapable analogy. I'd say that the key here (or the key of the q-derivative) is that you write f'(x) as (unitful stuff)*(unitless stuff). If x were measured in meters, we don't want our complicated math to have units in it, so we don't really want to write "x+h" because that forces h to have units of meters as well! Instead, we can formulate our problem as we vary the unitless parameter t. Then, we write our answer as (unitful stuff)*(unitless function of unitless ratios).
I'm bad at phrasing this stuff without using physics language, but it's the same trick you'd use if you're solving the nonlinear ODE L*f''(t)=-g*sin(f(t)) (a pendulum). Even though the problem is nonlinear, it's easy to prove that the period of the pendulum is T=sqrt(L/g)*r(f(0)), where r is a unitless function of the initial angle. If we really really insist that I can't use the word "units" or "meters" (say I'm doing a graduate math course homework problem!), then I can simply rephrase things in terms of rescaling the variables I work with, like choosing to work with t*x instead of h!
It also comes in handy when debugging numerical analysis code. But OK, I've ranted enough for one youtube comment, thanks for the stimulating video!!! :)
Yeah the unites aspect does make it somewhat elegant. Btw - new video is up finally. I'd love to hear what you think . Thanks! - HC ua-cam.com/video/aF5mTK4792s/v-deo.html
It's a really nice concept. I do have a comment on the graphics: when writing functions it's better to keep the notation "upwards" instead of italics - it makes it easier to separate variables from function names. In LaTeX specifically (which is what manim uses under the hood), most common functions have macros, e.g. \sin, \cos, \log, \ln, etc. If there isn't, one can always use \text{function}. For exponentials in the natural base, one can write e^{x}, but that again makes an italic e. So instead, I personally define a new command:
ewcommand{\eu}{\mathrm{e}}. Then \eu{x} is typeset correctly.
Thanks for the comment. I didn't use Manim, just KeyNote which has a LaTex editor. But I'll certainly play around with it.
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This definition is equivalent to the other (if the limit exists in both case), indeed every way of approching x can be either done by x+h or tx. Since the real topolgy is second countable, in particular every point have a countable open basis, we may choose countable sequence to fully caracterize limit (instead of nets) now you can show every sequence approching x can be done by some t_i that you mutliply with x or by some h_i that you add to x, taking i--> to infinity you get x. Supposing limit of (f(tx)-f(x))/(tx-x) exist then the classic derivative exists and equals the same, you can have problem where the denominator is undefined but you have to change. That's is why I think the first definition of derivative is the best way, in other word the linear approach is better.
actually, t^4-1 has the factors (t+1), (t-1) and (t^2+1), which can be further factored into (t^2+i) and (t^2-i) on the complex plane.
It would be (t + i) and (t - i). Also, factoring is only for rational numbers, and the complex plane does that fall under this requirement so it’s technically not a “factor.”
The beggining for another maths related channel? New sub!
Yes! I have whole series of videos in mind, especially for 4D stuff. I'm so happy this first one got off to a good start. Thanks!
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about a year ago i had to prove the derivative of √x and it took me hours, so when it showed up in the video, i was overwhelmed with emotions i hadn't felt in a while. university level maths is no joke, and i'm glad i'm just studying computer science with a little bit of math on the side.
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Zero is a tricky problem where there are infinitely many `1/0` values, but only the single zero covering all the subtly different uses of 'zero'. Given maths typically starts with the idea of 'next' (or increment), then yes, the 'h' method is nearest to that.
The 't' method works real well for regular folks who haven't yet reach the required level of pedanticism ;-)
😆
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This was very enjoyable. Disclaimer: I am neither a mathematician nor a teacher, so my opinion comes with no authority whatsoever.
Convergence is hard to teach/learn correctly, especially if a student will be exposed to higher mathematics further in their education. As convergence is defined in terms of distance rather than a ratio, I feel that the traditional approach might be better at building consistent intuition.
I'd probably still throw your method in here and there as an illustration of how things can be approached in different ways. It is always nice to see that alternative approaches give consistent results.
Also, well done on the manim scripting, the video flowed very smoothly.
Thanks very much! I think I see your point, though if we treat x as a fixed value, we're still doing the exact same thing, reducing the distance between the points to zero. You could directly substitute h = (t-1)x to change between definitions. Then the (fixed) value of x just changes the 'rate' at which h goes to zero. But yes, it's more of a way of demonstrating that any way of bringing the points together accomplishes the result of the derivative, even something like x^k and letting k--> 1. Actually I didn't use Manim, though I'm planning to learn to learn at some point. Just Apple Keynote for the slides/algebra and GeoGebra for the graphs.
@@HyperCubist "But yes, it's more of a way of demonstrating that any way of bringing the points together accomplishes the result of the derivative ..." - This is a really valuable point to learn early on, and I don't think it is taught explicitly in most places.
Learning Manim can be a bit of a pain if you are not familiar with programming already, but it is worth the struggle if your current toolset can't do what you need for future videos. I've obviously subscribed so I don't miss anything :)
@@michaeldamolsen Thanks for the sub. Not posting in a while has killed the momentum, not sure if even subscribers are seeing the new video. Pls check it out if you haven't, I'd love to hear what you think! Thanks! - HC ua-cam.com/video/aF5mTK4792s/v-deo.html
@@HyperCubist Thanks for the heads up on the new video! I'll watch this evening and comment on it to generate a bit of momentum :)
Differentiation made me sense to me when I realised it was a higher order function. For f, it gives you d(f, x) = my+c, that is the line/plane/surface that touches f at x when y=x. If you always set y=x, and you inline d, you recover high-school calc.
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This is great content. I hope to see more from you. Subscribed.
Count on it!
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sin(tx)-sin(x)
=2sin((t-1)x/2)cos((t+1)x/2).
We can factor a cos(x) out of the limit but this still requires the limit of sin(θ)/θ.
Similarly, for the exponential, we can factor e^x and we’re left with a numerator of e^((t-1)x)-1 which still basically requires the same method as the usual proof.
Yes - iirc I worked out that one the same way, but left it out as it wasn't as elegant as the others, and doesn't really offer much of an advantage conceptually or algebraically. Still cool that it works.
Can you explain the steps in more detail please.
@@doruk1713 which steps? Have you tried writing out the limit with sine and e^x to see what I am talking about?
@@GreenMeansGOF I've figured it out thank you
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Excellent video! Very refreshing!
You blew up my mind, and that thing made me comment.
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Michael Penn talked about this so called "Quantum dereivative".
I did see that! Though well after completing this video and having stumbled onto the q-derivative on Wikipedia.
Yeah I had come across this quite early on actually, I don't have a formal education in calculus but from the stuff I do know there have been multiple definitions of the derivative I have come across, including the symmetric derivative which is lim h->0 (f(x+h)-f(x-h))/(2h), I found that particular one interesting because it actually gives a value to some discontinuous functions at their discontinuity that make a lot of sense.
Yes! That was exactly the form that got me playing around with the definition and finding the tx form on my own, before learning it was already a thing :)
@@HyperCubist If I remember correctly when I was playing around with the symmetric derivative I also tried to come up with a symmetric version of the multiplicative version and don't remember if I succeeded or not but I do remember that one of the forms I tried was lim h->0 (f(hx)-f(-hx))/(2hx) and also found some variations of this for plugging in higher order polynomials that were interesting in properties.
@@christopheriman4921 I think a symmetric form would be something like lim [f(tx)-f(x/t)]/[tx - x/t] as t goes to 1.
@@HyperCubist That reminds me that I actually went from what I said there to lim h->0 (f(x(1+h))-f(x(1-h)))/(2hx) and also started plugging in higher orders of h into that to see what happened.
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Thanks for such vivid explanation.
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really love that presentation style thank you sir
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9:40
Regarding a proof of the identity x^n-1=(x-1)(x^(n-1)+...+1), you can also just note that (x^(n-1)+...+1) is a geometric sum and equals (x^n-1)/(x-1) which gives you the result immediately.
Also works!
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As an Engineer who has been exposed to a lot of math in my life, I find this video fascinating and I love this new approach.
That said, I don't think that this is how I would have liked to learn derivatives the first time I came across the concept. Delta(y)/Delta(x) which converts to (y1-y)/(x1-x) and from there to [f(x1)-f(x)]/(x1-x) defines the concept. And from there to call x1 = x+delta(x) there is only one step. But it would not even be my immediate step, I would stick with the "moving" point approaching the reference point for a while before moving to the delta approaching zero. For example, for y=x^2, dy/dx = lim |x1->x| (x1^2-x^2)/(x1-x), working with the difference of squares we get lim |x1->x| (x1+x)(x1-x)/(x1-x) = lim |x1->x| (x1+x) and now we can replace x1 with x giving (x+x) = 2x.
That's basically how I'm doing it, except I let x1 = tx, and take the limit as t--> 1. That's what allows the x's to pop outside of the limit, giving you the form right away (and the algebra is cleaner)
@@HyperCubist ... Again, your new approach is great, I love it. The algebra is cleaner indeed at least in many cases. Just that, in my opinion, the addition of the t tending to 1 (or of the delta x tending to 0 for the matter) is a "distraction" in the middle of trying to understand the concept. That's why what is called the "alternative definition" (with a "free" value of x approaching the reference value of x) is the one I like most just at the very beginning. And that is just my subjective opinion.
@@adb012 OK I get what you're saying now. Yes, there's certainly value in that, no question. It's always great to see the same concept done multiple ways. As I recall I first learned lim x-->a f(x)-f(a) / x-a, which is basically the "alternative definition" at a specific point (a). Then we went straight to the standard form. I didn't run into the more general "alternative form" until later. I've tutored calc countless times, and I always have to figure out which way the students class introduces it so I don't confuse them.
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Beautiful approach. Breaking down the process into a simpler method is what makes math so exciting.
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Love the video. I enjoyed it very much.
Just one minor nuisance.
While watching the video for the first time in a stop-motion fashion (i.e., stopping after every point and studying the formulas) I got confused at 20:13 where you convert 1/u * ln(1+u) to ln(1+u)^(1/u) with an animation that seems to indicate [ln(1+u)]^(1/u) since the first expression is log of (1+u) and the animation just moves the 1/u part right above the entire expression with out any other hints.
After a while I figured out that this is not what you intended but it would have been easier to follow if you could add some extra brackets, as in ln[(1+u)^(1/u)] or some other indicator.
Again, thanks for a good video.
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Maybe the derivative isn't usually defined in this way because this doesn't generalize to higher dimensions all that well? We usually require the convergence along all paths, not just "linear" ones (tx only hits multiples of x)
Btw, we can substitute h=(t-1)x = tx-x as the x is constant for each limit. If t approaches 1, h will approach 0 and we get our usual definition of the derivative.
So we have "If the new derivative exists, then so does the usual". For real x not equal to 0, the converse is also true with t=(h+x)/x. For vectors x, this substitution doesn't work anymore tho (as we cant divide by vectors).
Interesting point, very possible.
@@reeeeeplease1178 "For vectors x, this substitution doesn't work anymore tho (as we cant divide by vectors)." Only if you don't try hard enough. ;)
1/v⃗ = v⃗¯¹ = (~v⃗)/|v⃗|² = (~v)/(v⃗ * ~v⃗) (~v⃗ is kind of like a complex conjugate, though for several inputs is also an identity function.)
If you're curious how this way of multiplying and dividing by vectors works out, look up Clifford/Geometric Algebra.
@@angeldude101 You should don't need to invoke Graßmann algebras just for a basic generalization.
@@friedrichhayek4862 The important part is that _is_ possible to divide and multiply vectors with the right system.
Thank you! I finally understand how calculating a derivative works! It's real nice how you show multiple examples with both the conventionally standard method and your alternative! Watching this is like practising along almost.
Thanks for the nice comment! I debated with how many examples and comparisons to show. Glad you found it useful!
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Blew my mind! I'll be retaking calc 2 & 3 this year, so this will be really useful.
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A different perspective. İmpressive
20:59 After that... OK REALLY IMPRESSIVE
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this is real clever. lovely!
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I like both definitions, and I'd use them both.
Yes, they complement eachother well.
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Calculus was my fav back in university years oh my when we got introduced to the L’hopitals rule
I came up with similar unorthodox methods of computing derivatives using the limit definition and was I was rather scoffed off by my peers and shunned by the teachers, I expected them to be a bit more supportive and guide me thru the technicalities and making the process more rigourous. But oh well we dont have time for that when we're preparing for competitive exams🙅♀️
It's a shame creativity in math is undervalued.
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@@HyperCubist I had already watched it this week earlier ✨ and It did refresh my intuition of the exponential function, I love your approach to the typical "math concept explainer videos", the steps you take feel as the obvious next step and make it seem super natural. The cohesiveness and coherence of the videos is great. Awesome content 💫💫 would love to see more from you!
@@Axenvyy Thanks so much Gurshan!
Very cool! As someone that gets the standard form conceptually, while also being rubbed entirely the wrong way by it, having this other option makes me feel much less anxiety at the prospect of doing them but hand... Would be cool to see a similar alternative for integration or some other related topics
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Pretty nice video!
Minor reflection.
It seems like for the examples you essentially get to replace binomial expansion and factoring an h with doing polynomial division. You kind of hand waved the polynomial division. But that doesn’t actually seem simpler. It’s not exactly harder either. But I see little gain here.
Also, this totally breaks when x=0.
Finally, you might argue that the more difficult bit about derivatives is the concept of a limit and rules regarding it. That’s because it’s usually not very clearly defined at this level.
The sqrt, abs and log results were pretty nice.
Always nice to see multiple perspectives.
I mentioned the zero part at the end, but it does break down for certain types of functions (see pinned comment). The polynomial factoring can be learned by a simple pattern, or you can prove it with synthetic division, or just verify it by multiplying. What I should have mentioned is that calc students already know that factoring, as it's just a rearrangement of the sum of a finite geometric series. Compare a one variable factoring formula where you know every term to the (x+h) binomial form, with two variables, and where most terms are difficult to write for a general power n.
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I can see this working like a math experiment in a classroom-setting. First, you demonstrate that the secant approaches the tangent (I think Newton’s lemmas in his Principia work well for this step). And then you ask how that might play itself out mathematically, and let your students play around with different models. This multiplication definition might arise, even though I would the majority of solvers to figure out the addition version first.
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This is a really clever approach. Thanks for sharing it.
Thanks!
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Essentially if ii understood well, what just happened is that the 't' factor defines the form of the coefficient to the X^n-1 term while the x term(exponent carrier) defines the form of the main derivative. we could even render that more rigourous by studying all such forms gotten from the algebraic mouvements expressed by factor 't' if we define some set "C" for (coefficient ) that can be adjoint to some set "F" for (Forms); all such forms carried by the main exponent carrier x. This could help us meausure more rigourously advantages in speed and other factors such as cleaniness of steps that arise due to the two methods. Donot know if all what i said makes sense but if not, we here to talk and contribute. Those were my thoughts.
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great vid, i look forward to your next upload
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Beautiful and intuitive 👏🏾👏🏾👏🏾
Thank you!
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Yooo, this is super dope! I've never really thought about redefining the limit-definition of the derivative to make it lend itself better algebraically to certain functions. Absolutely eye-opening and interesting! Now I'm wondering about perhaps other ways of extending the definition... (E.g. instead of an additive or a multiplicative change in `x`, perhaps one considers a different change in x, such that the algebra works out. In general, you'd have `z = g(x, h)`, where `g(x,0)=x`, and `g` takes whatever form is convenient for what you're trying to differentiate).
Thanks so much! Right - there are lots of ways you can define it. I was messing with z = x^k, and letting k --> 1, but couldn't find any functions that benefited from it. Also you can let both points move towards a single point. A variation that sometimes shows up in exercises is using x-h and x+h, with 2h in the denominator. If you find anything cool let me know!
@@HyperCubist Could be an analogous way to redefine the Definition of integrals?
@@HyperCubist The x+h & x-h method can be separated out as the average of the right-hand & left-hand derivatives. If the limit exists, the left & right limits must be equal, so the average *is* that limit.
As shown when you did the u-substitution, any method that makes the 2 points approach in the limit should work. In fact, all these methods can be summarized as f'(x)=[f(g(x)-f(x)]/(g(x)-x), for some g(x) approaching x.
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Nice I Learned Again Derivatives
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For a new maths channel, you've certainly emerged swinging.
Yeah, I was quite astonished. Stay tuned for more!
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Feeding the algorithm 🥘! I just learned about the q derivative from another channel, so this was a cool follow up.
All praise the algorithm.
Unfortunately not posting in a while has displeased the algorithm. But thenNew video is up finally. I'd love to hear what you think . Thanks! - HC ua-cam.com/video/aF5mTK4792s/v-deo.html
Thank you and thanks again for the video! Dankie AS they say Afrikaans !
I loved it! A New Definition of the Derivative?
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Wow fabulosa definición para introducir los temas
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amazing video!! definitely deserves more views
Thanks so much! Be on the look out for more videos :)
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The reason it's not taught is really simple. Mathematician love generality, but the new definition only works for Real functions. In multivariable calculus, all paths have to have the same limit point, but with f(tx)-f(x) there is only one path possible: along the direction x points to (the existence of partial or directional derivatives does not guarantee a total derivative). In multi.dim. it has to be possible to independently choose a point and a direction and now it's the typical derivative or a directional derivative in the 1D case, since the only positive oriented normed basis vector is 1 itself.
You're right, it doesn't generalize to multivariable. The point is that it can be used as a pedagogical tool for learning power rule on a deeper level, and simplifies beautifully for several types of functions.
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I was taught a different way to calculate differentiate functions using h terms and you take O(.) of the terms to find the derivative. I don't know what this is called but I can find a reference if anyone is interested.
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Very well-put together! Personally I'm not entirely convinced of its use---to me, a lot of the nontrivial examples end up needing tricks or arguments that are roughly equivalent in trickiness/difficulty as in the original method's proofs anyway :)) So I don't really see any advantages---through intuition or ease of rigor---that this alternative definition provides. But I can still appreciate a video with clear explanations and great visuals, and this video is certainly excellent! The video was paced quite well and you have a voice well-suited for voiceovers.
Some particular comments I have:
-Re: |x|. Gonna start with a pet peeve of mine---people not liking piecewise functions and somehow thinking "closed forms" are superior. I would prefer to disabuse people of this notion as soon as possible, actually!!! Is there any particular reason for |x|/x to be preferable over the +/-1 piecewise? I'd argue no, and In fact, I'd argue it's worse, because it's just needlessly obfuscating the simple actual intuitive thing going on (-1 slope when negative, +1 slope when positive). Justice for if statements!! Elementary closed forms are tooooootally overrated, and I don't want to encourage my students to put them on even more of a pedestal :))
Also, computing the derivative of |x| using the standard definition is not at all hard. The only technical bit is needing to argue that when x!=0, we know that |x+t| and |x| have the same sign when t is sufficiently small----an argument that your alternate solution also still basically had to make, by the way, when you argued that |t| = t as t -> 1 :PP
-Re: sqrt(x). Again, your argument is of equal complexity to just using the standard definition of the derivative anyway. You still had to use the "conjugate trick" in the end to ultimately resolve your limit expression (while the substitution trick is clever, that's... just doing the conjugate trick, but wearing a fake mustache).
- Re: Power rule. I really don't see what's so unintuitive with the "binomial expansion". In fact, the argument here is the beating heart of understanding calculus intuitively: recognizing that (x+h)^n = x^n + nx^(n-1) h + (a bunch of other terms that aren't important because they go to 0). In fact, it feels disingenuously "over-spooky" to say that it _needs_ the binomial expansion and to write out all the binomial coefficients (you even expanded C(n, 2) into n(n-1)/2, what is the point of that other than to scare people??), when most of the coefficients are actually completely irrelevant and don't need to be written out.
(What follows is a somewhat weird take I have which is off-topic from general Calc 1 pedagogy)
Also: you can prove the power rule for rational exponents i.e. x^(a/b) by proving x^a and x^(1/b) separately and then invoking the chain rule. And then it's not immediately obvious how to extend this to real numbers... which I like! Because it forces students to think critically about what the heck something like 2^pi is _even supposed to mean_!!! I hate how much weirdness about real numbers just gets swept under the rug in math classes, since many students (my past self included) end up with the feeling that real numbers are totally normal things, when actually they are Incredibly Weird and that rigorously defining things with them is Incredibly Tricky
And yes, I'm talking about rigor here. Because if we disregard rigor, I'm pretty sure most people would already be convinced that the power rule holds in general, because it holds for integers and rational exponents, and like---polynomials are nice, so i wouldn't expect them to be too weird and suddenly break. So the only time (imho) we're ever actually discussing details is when in search of rigorous definitions and proofs. My understanding of the typical proof of power rule for real-number exponents is to let y = x^t imply ln(y) = t * ln(x) and then do implicit differentiation---which isn't so tough at all! Which finally leads me to...
- Re: ln(x). In many calculus classes, ln and exp are held off until the integral lesson, and then it is _defined_ to be the antiderivative of 1/x and e^x is defined as its inverse (from which many familiar properties of it and exp(x) can be derived). Once again, I emphasize that you somewhat _need_ to do this because otherwise there is this gigantic gaping hole of what a real-number exponent _even means at all_. Your proposed solution here takes a shortcut by just taking for granted that ln(x) and exp(x) exist and are things we can use... when they really haven't been defined yet at this point in time, and to be frank "I have no idea what they mean" :P
My preferred intuition is to define exp(x) by its taylor series, but obviously that can't be used until Calc 2.
I suppose you could go define e.g. 2^pi as the limit of the sequence 2^3, 2^3.1, 2^3.14, 2^3.141, 2^3.1415... but that involves a lot of handwaving about continuity that makes the argument feel weaker and more wishy-washy (freshman me sure wouldn't have been ready to deal with Cauchy sequences back then!!!)
Wow, thanks for the kind words and thoughtful reply! I won't hit every point, but... The big advantage (as I see it) is that (for these examples - power functions, logs, and |x|) all the factors of x 'pop out' right away before having to simplify the limit - so with almost no work you get to see what the derivative looks like to within a constant. Then you only have to solve a limit, with one variable rather than two. I emphasize this more in the follow up video (on my channel), but it's also a question of understanding intuitively WHY power rule reduces power functions by one degree, which can be seen for any real power, not just integers/rationals. So even if a student can't fully prove power rule for x^pi right from the get-go, they can understand exactly why it's proportional to x^(pi-1), using first principles (rise over run) - no chain rule or implicit diff to act as a barrier to understanding. The same logic applies to ln(x) - you can see exactly why rise/run is proportional to 1/x, and watch as it pops out of the limit.
As for if it's actually easier, try this. Use the standard definition ONLY (first principles - no chain rule) to find the derivative of x^5/7. I tried it myself the other day, and you CAN do it, but like the conjugate, you have to pull a factor expression out of thin air, and it's a god awful mess, nothing I would expect a student to do. Now try it with the 'tx' form. One substitution later and you're factoring simple polynomials (u^n - 1) that follow a simple and easily learnable form - one that students have technically already learned with finite geometric series. Then try the general rational case: x^m/n, or even x^-m/n, using both definitions.
Re: the conjugate trick - standard and useful as it may be - requires supplying factors from nothing, as in the example above. The substitution trick doesn't, you just have to factor what's already in front of you.
As for the pedagogical stuff and real powers, all I know is that students have been working with e^x and ln(x) since pre-calc, and treating them as continuous functions, so they must have some level of comfort with real powers.
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Some comments:
* When deriving the derivative of |x|, I think you did use the piecewise definition when you said as t->1, |t| is t.
* For the ln(x) derivative, you went from needing the chain rule to needing to know when to interchange logs and limits as well as knowing how to expand the series (1+1/n)^n.
You can actually work out a closed form for the abs value of x using the original definition.
Just sub in |x| = √(x²) and you can do it.
True - but that does use chain rule rather than going straight from the definition.
@@HyperCubist although the new definition you found does make it a lot easier to work out. I just figured out how to work it out using the limit definition while I was evaluating some functions that made use of the absolute value.
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Wonderful video! Incredible cross over of maths and satire. I haven't laughed so hard in ages. The slow burn of each example had me rolling! Keep it up!
Thanks ... I guess? I'm used to explaining to students and didn't want to rush through things, not knowing exactly what the audience would be.
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@@HyperCubist cool graphics. Maybe a follow up video could focus on how we get e from the natural logarithm
This just goes to show that school was never meant to teach you.
If this definition was introduced to my calculus class, so many more students would have understood
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Amazing video!, so comprehensive
Thank you!
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Great video!
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Thanks for spreading this method bruh. 🤪👌
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Thank you for the video!
I like your new definition. Well, I like all math.
I was exited to see how the h->1 defination would work for e^x. In the end, still a nice video. Good graphics and examples.
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Really elagant, I like it!
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I suppose this would be a good way to re-introduce limits in multivariable calc when considering limits along different paths. As far as intro Calc, doubtful it will really make things any simpler or easier to grasp.
this definition seems great for functions which contain multiplication
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I've derived a similar closed form version of the derivative of absolute value of x by using abs(x) = sqrt(x^2) which is the vector way of writing it lol. Interestingly, you get the fraction the other way around, as x/|x| instead.
Right - I came across that when playing around with it as well. It does use chain rule though - this way uses the actual definition of the derivative directly.
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Amazing…looks this may open up a new horizon?
As a challenge problem - try using this definition to prove the power rule for rational exponents: f(x) = x^(m/n). It's something the standard definition doesn't do easily.
Got close but I struggled with the lim t -> 1 (t^m/n - 1)/t - 1, any hints? I know this should come out to m/n but I'm not entirely sure how.
@@squeezy8414 You're close - check out the substitution I did in the square root example, and the factoring of x^n - 1 in the power rule example.
@@HyperCubist Got it, tried a u = t^m/n substitution first but that didn't work but then I did u = t^1/n and that gave both the numerator and denominator integer powers - thanks for the tip!
Now try it for negative powers: x^-n, or better yet x^-m/n
Have you seen John Gabriel's 'New Calculus'? I like his as well but it is radically different as it doesn't use limits at all. The definition is purely algebraic as it relies on the exact definition of the tangent line. It essentially combines left and right handed derivation into one.
Great video. Thank you
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Very nice video, I would have really enjoyed a proof that the two are equivalent, I think it is enough to say that it has the product rule and the derivative of x is 1 (because then it is easy to have the derivative on all polynomials, including Taylo approximations of C^2 functions).
It was already 23 minutes long lol! But they are equivalent except at x = 0, in which case there's the lim x-->0 path. And even then there are some weird exceptions (see the note in the link). But you can prove all the standard rules in similar ways as the x+h def. Keep in mind it's meant as an elegant tweak, not a replacement.
A formal proof for all x+h there is an equivalent value of t such that x+h=tx?
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It's quite nice that t is unitless.
I hadn't really thought about that, but yes it is!
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I was hoping for you to mention that this method is like saying f'(x) = g(x).f'(1) Where g(x) is generally easier to find in shown cases, and f'(1) is some limit we'd have to solve 😅
You know someone else made that same comment, and I was scratching my head trying to figure out why that's interesting? I mean you get a function of x, times a limit, which is constant. What's the significance that it happens to be f'(1)? I could see that being interesting for f(x) = a^x using the old definition, as f'(x) = f(x)*f'(0); it's proportional to the original function. But in this case the derivative is proportional to a different function, so what's the significance? Help me out - I'm legit curious.
This is brilliant
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A reason why I think this definition is used less is that it doesn't generalize well to multiple dimensions. As soon as you have multiple dimensions your limit needs to be able to approach for every direction. In 1d this is fine since q can fluctuate as q1. In multiple dimensions you would need to define some multiplication on vectors which can be quite tricky.
Right - I'm not advocating for it to replace the standard, just as an alternative that simplifies beautifully for some functions in 1D Calc.
It's completely possible to define a product on vectors. The best version that I've seen has been to define a bilinear associative product such that a vector squared is a scalar. However this does mean the result is a noticeably different kind of object than you started with, and also that it's non-commutative in general, so it would be interesting to see how this version would work with such a product.
Then again, wouldn't the traditional definition also need to divide by a vector if h is meant to approach 0 from all directions?
@angeldude101 For the traditional derivative you can extend it with the total derivative, which in my opinion is a very natural extension.
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I think if you use sum (difference) to product formulas, you can get the derivatives of sine and cosine.
But you would still need to know sin(u)/u -> 1 as u ->0.
If you're willing to use implicit differentiation and the arc length formula, you can derive the derivatives of sine and cosine in terms of each other without proving that sin(u)/u goes to 1 as u goes to 0.
Could you please explain how
@@doruk1713 I don't know if you were asking me as I think @matthuelson3409 had the more interesting comment. But if you were, here it is (without carrying the limits):
sin(A) - sin(B) = 2sin([A-B]/2)cos([A+B]/2)
so
[sin(tx) - sin(x)]/(tx-x) = 2sin([tx-x]/2)cos([tx+x]/2)/(tx-x)
= cos([tx+x]/2)*sin([tx-x]/2)/[(tx-x)/2]
-> cos(2x/2)* sin(u)/u -> cos(x)
cos(A) - cos(B) = -2sin([A+B]/2)sin([A-B]/2)
so
[cos(tx) - cos(x)]/(tx-x) = -2sin([tx+x]/2)sin([tx-x]/2)/(tx-x)
= -sin([tx+x]/2)*sin([tx-x]/2)/[(tx-x)/2]
-> -sin(2x/2)* sin(u)/u -> -sin(x)
The quick and dirty way: Use x^2 + y^2 = 1 and (x')^2 + (y')^2 =1 (this is parametrization by arc length). Also, we use initial conditions x(0)=1 and y'(0)>0, so the curve is traced counterclockwise.
Take derivatives of x^2 + y^2 = 1, implicitly: 2x x'+2y y' = 0, implying y y' = -x x' and squaring this yields y^2 y'^2 =x^2 x'^2.
Now, multiply both sides of (x')^2 + (y')^2 =1 by x^2, obtaining x^2 x'^2 +x^2 y'^2 = x^2. Replace x^2 x'^2 with y^2 y'^2, obtaining y^2 y'^2 +x^2 y'^2 = x^2. Factor out a y'^2 on the left: (y^2 + x^2) y'^2 = x^2, and, since y^2 +x^2 =1, we arrive at y'^2 = x^2. The initial conditions then yield y' = x. Interpret y = sin t and x = cos t (we are on the unit circle, after all), and we're home free. There are subtleties to be ironed out, mostly around continuity and taking square roots properly.
@@iambic-kilometer It is very explanatory, thank you 🙏
Maan I'm flabbergasted ❤
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Very nice video. However, correct me if I'm wrong, this approach would not be easily extended to the multidimensional case.
You’re right
Yes, it doesn't extend in the general case. Just another tool in the toolbox for certain 1D functions, great for building up intuition. Btw - New video is up finally. I'd love to hear what you think . Thanks! - HC ua-cam.com/video/aF5mTK4792s/v-deo.html
I thought about something like one day, but never push it, and I'm glad you did because it's infact interresting ! To me, both definitions are good, it's just a matter of "does the function I want to know the derivative of behave more nicely with addition or multiplication ?" ; of course power functions behave much better with multiplication since for them f(x*y) = f(x)*f(y), and exponentials behave much better with addition since f(x + y) = f(x)*f(y).
I'm wondering now if one could take even another definition for "z", adapting this definition to be the best suited for the function considered ; for example if you want to know the derivative of a function "f" for which f(x^y) = f(y)/f(y) , could we create the definition of derivative better suited for the calculation ?...
If any ideas please share ^^
And really cool video :)
Thanks! I’ve been looking for other variations, like z = x^k, but haven’t seen any functions that work well with it. Please share if you find any!
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Much more intuitive and easier.
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