derivatives "defined" without limits?!?

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[@nikitasstathatos7970]

Another issue is that it's a bit awkard to talk about tangents of linear functions. If we take f(x)=cx, then the tangent to the graph at any point coincides with the function, so there's infinite intersection points instead of a unique one, which means that we cannot calculate the most basic of derivatives! Seems pretty similar to the inflection point case, so perhaps it also has to be excluded.

[@coc235]

In fact, by his definition, literally every number besides the actual derivative is a "derivative" of a straight line, since these lines will have exactly 1 intersection point.

[@hooked4215]

@@coc235 Any straight line has exactly two intersection improper opposed points.

[@ianfowler9340]

I never thought of that. If you apply the revised method to y = mx + b at P(p,q) on the line, then the equation of the tangent is the line itself. y - (mp + b) = m(x - p). Nice.

[@hooked4215]

@@ianfowler9340 Somebody has to get out more often.

[@blackyusylvean]

For the exponential case, it is doable using a non-elementary function called the Lambert W function, which is defined as the inverse of the function xe^x. Since said function is not bijective, the Lambert W function has infinitely many branches, however since we're looking to solve e^x = m(x-1)+1 in the *reals*, only the two real branches W[0] and W[-1] will work. W[0](x) is defined for x in the interval (-1/e ; +inf) and W[-1] is defined for x in the interval [-1/e ; 0) Therefore an equation involving the lambert W only has one solution if its argument is positive. Solving e^x = m(x-1)+1 for x using the Lambert W: exp(x) = m(x-1) + 1 -> exp(x) = mx - m + 1 -> 1 = (mx - m + 1)exp(-x) -> -1/m = (-x + 1 - 1/m)exp(-x) -> -1/m exp(1 - 1/m) = (- x + 1 - 1/m)exp(-x + 1 - 1/m) We have something of the form w = z exp(z), therefore solutions are all of the form z = lambertW[k](w). In our case, we have - x + 1 - 1/m = lambertW(-1/m exp(1 - 1/m)) -> x = (1 - 1/m) - lambertW(-1/m exp(1 - 1/m)) As said before, in order to have only one possible solution, the argument of the lambertW must be positive so as to only be defined on the 0-branch. Hence we must have -1/m exp(1 - 1/m) >= 0, which is only true if m is strictly negative. In fact, any negative m will do, so this does not work very well since we expected to get at least a positive slope. The only m such that there is exactly one solution is m=0, in which case we have exp(x)=1 -> x=0, but that still isn't the value we were expecting (m=1)

[@EricOfYork]

Minor arithmetic error at 9:10 the constant term of the quadratic should be m^2-2m+1, not just m^2+1 This leads to a simplified discriminant of 4m^2 -4m + 1, which factors to (2m-1)^2

[@kappascopezz5122]

9:25 The quadratic formula only works when m =/= 0, so you first have to check if m=0 is an option. And indeed, when you enter m=0, you get 1-x=0, which has exactly one solution. I think one way to exclude these "bad" cases of m is to add that within a neighbourhood of x=a, the tangent line should either be all less/equal or all more/equal than the function in question.

[@hm4tt0s]

Indeed, furthermore I think if that distinction is not made, just looking at the graph of sqrt(x), every m

[@guilarai]

The function can also be above the tangent line at one side and below at the other side (like x^3 at x=0)

[@saroshadenwalla398]

​​@@hm4tt0sBut if m is not 0 then the quadratic formula can be applied and you only get x=1/2. So I don't see how an m

[@saroshadenwalla398]

​@@guilaraix^3 at 0 would be eliminated for having an inflection point there.

[@kantanlabs3859]

When I was young (in the sixties in France) the ideas of continuity and derivability were defined relying on the idea of bounded values for arbitrarily small intervals. Theses notions were tackled from a somehow topological point of view (compacity, open of closed sets…). I just remember that demonstrating that a function is derivable in a given interval or even a single point was much tedious than it is today with the modern formalism relying on limits and associated theorems.

[@knedl9796]

I'm not sure whether this definition generalises that well, since, for example, sin x = mx only has one solution for many different negative values of m. But I think this following definition could work: First, define A := {all real values m for which m(x - a) + a - f(x) >= 0 on the interval (a - ε, a], for all small enough values of ε > 0}. This set should be of the form (-∞, c) or the form (-∞, c], for some real number c. This is because if some number is in the set, then all smaller numbers are in the set as well, since the value of m(x - a) increases when m decreases. Likewise, if a number is not in this set, all the bigger numbers also aren't a part of this set. Now you define the left derivative of the function f at a as the value c. Similarly, we can define the right derivative. The set B := {all real values m for which m(x - a) + a - f(x) >= 0 on the interval [a, a + ε), for all small enough values of ε > 0} must be of the form (d, ∞) or [d, ∞), for some real number d. The argument is similar to the one above. Now we define the right derivative as the value d. Finally, we define the derivative to be the values of the left and the right derivative, in the case that these two are equal. Otherwise, no derivative exists. Your condition did not work in inflection points, because the function m(x - a) + a - f(x) changes sign for all values of m, as opposed to sign staying equal for some m (when m is the derivative) in other functions. But by separating derivatives into the left and the right derivative, we can just treat the two sides individually. We can then focus more on the sign of the function instead of the number of solutions, which brings us to this defintion.

[@robert.ehrlich8942]

Your guess at 14:36 that if you are away from an inflection point is false. A good counter-example for this is the function f defined by f(x) =x²sin(1/x) for x not equal to 0 and f(0) = 0. This function has a derivative equal to 0 for x=0, the value of your "m" should be 0, but the equation f(x) = 0 has an infinity of solutions in any interval containing 0. It can be argued if you are really "away from an inflection point" at x=0 since there is an infinity of inflection points in any interval containing 0 (the sames as above) but at x=0 we don't have an inflection point.

[@ivandebiasi6657]

I'm missing something or either in the case of f(x) = sqrt(x) and f(x) = sin(x) any value m

[@guilarai]

For sqrt, if you consider negative values of m, squaring both sides always creates an extraneous solution. So basically, imposing the uniqueness of the solution in the quadratic equation in terms of x forces m to be non-negative. Michael was a bit quick to assume the uniqueness of the solution in sqrt(x)=m(x-1)+1 implies the uniqueness of the solution after squaring both sides... or blatantly sweeped that under the rug 😂 For sin, I guess the m>1 values already made the point. (and just for the sake of accuracy, m would have to be negative enough for the tangent not to cross the other waves)

[@ianfowler9340]

Here's another nifty extension. Find the equation of the tangent to y = x^3 - 3x^2 - 3 at the point of inflection. Then intersect the tangent with the cubic. You will get an equation that has a triple root at the x-value of the inflection point. Works in general. Nice proof.

[@MK-13337]

For the quadratic (and the square root) case, you are forming an equation that essentially asks wether you only have one *global* solution. Your definition of "near" seems off. If I'm allowed to be arbitrary with what is considered "near", then I think all nice functions seem to have all or most real values as their "derivative", and we are cherry picking the known value and going "aha! If our solution set is most of the real line it includes the actual answer and thus is useful somehow"

[@ianfowler9340]

Another way to force the double root at the point of contact where x = a is to divide f(x) by (x - a)^2 and inspect the reminder. Which must be linear since we are dividing by a quadratic. x^2 divided by (x - 3)^2 gives a quotient of 1 and a remainder if 6x - 9, the tangent equation. A cubic example: Find the equation of the tangent to y = x^3 - 2x + 3 @(1,2) Divide x^3 - 2x + 3 by (x - 1)^2 = x^2 - 2x + 1 You will get a quatient of x + 2 and a remainder of x+1 so the tangent is : y = x + 1.

[@edwardlulofs444]

Ohh. I think this is my favorite video of the last 12 months! I love the math where you might think some procedure is correct and simpler than a more complicated definition - but it’s not correct. It reminds me of a math lecture in 1970 where the professor showed that irrational numbers can’t be used as a base. That is based on 2 works, based on 10 works. But base pi does not work. Excellent video! This made my day.

[@miraj2264]

At 7:41, probably easier to let y = sqrt(x). Then you have y = my^2 - m + 1 ==> my^2 - y + (1-m) = 0. We want the discriminant = 0 so we need b^2 = 4ac ==> 1 = 4m(1-m) ==> 4m^2 - 4m + 1 = 0 ==> (2m-1)^2 = 0 ==> m = .5.

[@spicymickfool]

Descartes found derivatives via osculating circles. The radius of a tangent circle is perpendicular to the tangent line. Alternatively, find the intersection of $y=mx+b$ with a parabola. Select m and b so that the discriminant is 0. The result gives you the tangent line the slope of which is 2Ax+B as expected.

[@GabrielOliveira-ql2cg]

I think there is something missing, no? For the exponential case, for exemple, at any point (a,e^a), e^x=m(x-a)+e^a will have a unique solution for any m

[@jimbobago]

10:17 I get 8m^3, not 4m^3 but then 1/2 isn't a root.

[@PROtoss987]

John Gabriel's 'New Calculus' was an attempt at this with the intention of never going to limits, although I'm pretty sure there's no mathematical rigour to be found in it

[@benardolivier6624]

Inserting m=1/2 into 4m^3-4m²+4m-1 doesn't give zero, there must be an error in this result... (the factorization below gives 4m^3-6m²+4m-1)

[@JustinWilsonPhysics]

Looks like there is an error in the quadratic equation for x, c = (m-1)^2, not c=m^2 +1.

[@minamagdy4126]

The error I saw is that c =-m^2+1 = 1-m^2, not 1+m^2

[@Apollorion]

@@JustinWilsonPhysics That c = (m-1)^2 is indeed correct and leads to the discriminant becomes equal to (2mm -2m +1)^2 - 4mm(mm - 2m + 1) = 4m^4 +4mm +1 -4m +4mm -8m^3 -4m^4 +8m^3 +4mm = 4mm-4m+1 = (2m-1)^2 . So this discriminant being equal zero implies just the single solution m=1/2 .

[@MichaelFJ1969]

I've seen the derivative (or rather the derivation operator) *defined* from the product rule (Leibniz' rule): (fg)' = f'g + fg' together with simple linearity: (af+bg)' = af' + bg' for constants a and b. These two properties are (as far as I recall) enough to derive (pun intended) all other properties.

[@pietergeerkens6324]

The factoring at 11:12 is incorrect. (2m - 1) (2m^2 - 2m + 1) = 4m^3 - 6m^2 - 1 and not the required 4m^3 - 4m^2 + 4m - 1. Likewise by substitution of 1/2, the value for the cubic at 1/2 is 1/2 - 1 + 2 - 1 = 1/2 0. This seems to follow from the earlier error at 9:10 for the constant term of the cubic.

[@Steindium]

I've looked into this "definition" before, and you could solve the e^x one, if you can somehow prove xe^x = c only has one solution at x = -1 (its minimal point) without using derivatives: e^x = m(x - a) + e^a -> e^(x - a) = me^(-a) (x - a) + 1 -> 1 = (me^(-a) (x - a) + 1) e^(a - x) -> -e^a/m = ((a - x) - e^a/m) e^(a - x) -> -e^a/m • e^(-e^a/m) = ((a - x) - e^a/m) e^((a - x) - e^a/m) only has 1 sol if -e^a/m = -1, or m = e^a. One thing I will add though, I'm particularly interested in whether or not you can then prove the Fundamental Theorem of Calculus with this definition as well (you'd probably have to define integrals without Riemann Sums, or any other definition that requires limits)

[@Miparwo]

Far more elegant and general is to use nilpotent units, for automatic differentiation.

[@hbowman108]

The most streamlined way to algebraically differentiate an algebraic variety at a point is to linearize it. You express the variety as the intersection of the zeros of multivariable polynomials and change variables. For instance, if we have y²=x at (1,1), we replace x with u+1 and y with v+1. Then we have v²+2v+1=u+1. So, -v²+u+2v is zero at (u,v)=0, and then we get the cotangent from the linear approximation u+2v=0, giving us that (1, 2) is in the cotangent space. Then it is a simple matter to find the vector perpendicular to the cotangent space. The slope of (1,2) is 2, so the derivative is -½. If the dimension of the cotangent space is "wrong", you're at a singular point on the variety.

[@peterg76yt]

I'm not clear how you are defining 'tangent' without it just being a limit in disguise.

[@ianfowler9340]

When I was in my senior year in high school one of my 3 courses dealt with the conic sections. We learned this method which, because the curve was restricted to the conics, always reduced to an intercection equation which was quadratic. Hence discriminant = 0. Try the ellipse : x^2 + 4y^2 = 8 at the point P(2,1). Force D = 0 to get (2m + 1)^2 = 0. Very cool. But notice that this value of m forces the intersection equation to have a double root at x = 3 when you sub in m = 6. i.e. (x - 3)^2 = 0. This is actually Descartes' Revised Method by forcing a double root at the intercection of the curve and the tangent. I think it was only intended for "polynomial" functions. But the double root method does work for the Folium of Descartes : x^3 + y^3 -4xy = 0 at P(2,2). So Descartes' Method reduces to discimant = 0 in the special case where the curve is a quadratic (i.e. a conic section).

[@pwmiles56]

At any point of an algebraic curve, you can make a tangent line in homogeneous coordinates. E.g. write your example f(x,y,u) = x^3 + y^3 - 4xyu = 0 where u is an extra coordinate introduced to make the equation homogeneous. The "actual" x and y are x/u, y/u. The coordinates of the tangent line are (m1, m2, m3) = (df/dx, df/dy, df/du) in which partial derivatives are understood I.e. (m1, m2 m3) = (3x^2 - 4yu, 3y^2 - 4xu, -4xy) The line equation is m1 x + m2 y + m3 u = 0 Converting back to inhomogeneous coordinates (m1/m3)x + (m2/m3)y + 1 = 0 With a bit more work you can show that, if (x,y) is indeed a point of the curve, the tangent made in this way yields a double root (or higher) for the intersection.

[@ianfowler9340]

@@pwmiles56 I did not know or consider that. Very nice. Thanks for taking the time to reply.

[@mathunt1130]

I believe the string theorist Robert Diigraaf(sp) did talk where he started out with the well known derivative definition and presented all the generalisations. I believe that the one of the examples he uses is the one here The square root example may have been easily done, by simply considering the equation as an equation in \sqrt{x}. The calculation is far easier.

[@eduardochappa4761]

Michael, the idea that the tangent line only intersects the graph of a function at one point locally can be associated to the idea of concavity, which in terms of second derivative means that the graph of the function is on one side of the tangent (locally). What you did in your video is to put the tangent line of the graph fully on one side of the tangent line (for example that is what you do when you say that the discriminant of a quadratic must be equal to zero), but in reality if you are looking to find one line that intersects the graph only once near the point of tangency, pretty much any line will do. For example, for f(x) = x^2 the line y = 2x intersects the graph of f(x) at 0, and nowhere close (for example, there are no other intersections in the interval (-1, 1)). For the type of functions you looked at and the definition you used, pretty much any line is tangent. On the other hand this definition has the issue of how to distinguish between the differentiability of f(x) = x^3 at 0 (in the normal sense) and the lack of differentiability at 0 of g(x) = |x|. This definition does not seem the make any difference in the definition of derivative. To me it seems that this is an attempt at trying to define something that is definitely not related to tangency. There is no requirement that a tangent line must intersect a curve only at the point of tangency and nowhere else close by. In fact, you can find functions f(x) that have derivative at a point "a" but their tangent line intersects the graph of f infinite number of times in any neighborhood of "a". As a hint of how this could happen, notice that if the second derivative of f at "a" is not zero, then the tangent line at "a" intersects the graph of f only at "a" near "a". I would not have published such paper, had I been the editor of that journal. It seems bad.

[@EMAngel2718]

I wonder how much stronger you could make this "definition" by stating that it has to be at an edge of any range of numbers that satisfy the condition In the case of sine it would at least reduce it from infinitely many false solutions to just 1 Edit: Actually thinking on it further the one remaining false solution would be excluded cuz it's at the edge but not in the range. This all said I do kinda wonder if it would be possible to prove a number is not in the bulk of a range without limits

[@JacobWakem]

Another way is to just treat x as a time variable and ask for the rate of change of f at x.

[@zuzaaa1998]

This notion of the derivative is closely related to the concept of subderivative in convex analysis

[@jay_sensz]

Finding the derivative of an arbitrary polynomial at the point x=a without using limits is pretty straightforward by replacing `x` with `(x-a)+a` and multiplying everything out, resulting in the Taylor polynomial at x=a where the linear coefficient is of course the derivative. In the case of x^2, we get ((x-a)+a)^2 = (x-a)^2 + 2a*(x-a) + a^2, so the derivative is 2a.

[@byronwatkins2565]

You could probably make this work for inflection points by adding the requirement that the curvature is zero. You could probably use the McLaurin or Taylor equivalent of trancendental functions. But, justifying this without calculus is problematic.

[@coreybrady6305]

Hi - Thank you for your channel, which is incredibly thought provoking. In this particular post, I think that the goal as posed might be off-base. In your f(x)=e^x case, m=0 (and many other values) will work and can be nicely verified from the monotonic-increasing property of the exp. (By the same reasoning we could find that odd-power monomials also will have non-tagnent horizontal lines that intersect only once.). Rather I think that the really exciting idea that is behind the algebra in THIS post might actually point to something interesting and special about the quadratic (y=x^2 or x=y^2). Parabolas are amazing shapes with very cool properties, including in differentiation. [For example, taking a symmetric definition of derivative (secants that intersect at a distance h/2 on both sides of the the focal x=a point), gives you an h-independent slope in the quadratic case. This is because of the "merton rule" of average/midpoint velocity). Super interesting as a special feature of quadratics.]

[@maurobraunstein9497]

I wonder if you can do something with Taylor series. I'll give an example because it's easier than generalizing: consider f(x) = x^2 at x = 3. This can be written as 9 + 6(x - 3) + (x - 3)^2, right? For any polynomial p(x) (finitely many terms, non-negative integer powers), you can write it as p(a) + p'(a)(x - a) + p''(a)(x - a)^2/2 + ..., so if you do this, you can just read off the derivative as the coefficient of the (x - a) term. This means that the derivative of f(x) = x^2 at x = 3 is 6, since that's the coefficient of (x - 3) in the expansion. Using this, you can trivially prove linearity, for example, and that if p(x) = x^n, then p'(x) = n·x^(n - 1). I'm not sure how we would extend this to functions that are not polynomials, because we can't really use infinite series without a notion of a limit. Maybe there's a way. I think we can talk about formal infinite series, without requiring convergence, but I'm not sure how to connect them to functions. We can also derive the product rule and chain rule for polynomials, posit that the derivative of a non-polynomial follows the same rule, and use implicit differentiation to find derivatives for fractional and negative powers, which will, of course, still just be n·x^(n - 1).

[@plislegalineu3005]

In Poland we use the uppercase letter delta (that triangle) for the discriminant and we use it a lot cuz when we learn the quadratic formula (which doesn't have a name here) we first calculate the discriminant and then plug it into under the square root

[@VideoFusco]

The real way to define derivatives without limits is via infinitesimals and the non-standard analisys.

[@PawelS_77]

I think using the word 'near' implies taking a limit in general case.

[@niom-nx7kb]

just curious, what does this "new" way of finding the derivative contribute to?

[@edmundwoolliams1240]

Fun. As Feynman said: "Physics is like sex. Sure, it produces practical results, but that's not the reason we do it!"

[@ianfowler9340]

Historically, it gave a method of finding the slope of a tangent in an era of pre-calculus. You can also find the equation of the tangent if you divide x^2 by (x-3)^2 and inspect the remainder : 6x+9 . This works for all polynomials.

[@niom-nx7kb]

@@edmundwoolliams1240 I see

[@edal7066]

it has to do with algebraisation of the analysis as epsilon-delta language becomes, in this case, obsolete. It leads to algebraic expressions that, in general, better implement algorithmically.

[@udic01]

10:12 you have an error with the -4x^2.

[@59de44955ebd]

If I'm not mistaken, there is another bug at the very end of the video, instead of e^x = m (x - 1) + 1 it should be e^x = m (x - 0) + 1, or just e^x = mx + 1.

[@59de44955ebd]

I guess like many others I realized that 1/2 no way can be a solution of 4m^3 - 4m^2 + 4m - 1 = 0, since only the first term would result in a non-integer (1/2), while all other terms would be integers. Fortunately @EricOfYork found the bug.

[@annaclarafenyo8185]

This is nonsense as a definition. Consider f_k(x) = exp(-1/x^2)*(sin(kx)) and convergent sums of this type of thing.

[@barteqw]

First time i met derivative without any limit is so called q-derivative. It's used in quantum calculus, combinatorics and machine learning.

[@pwmiles56]

How about f(x)=x^3 ? It has an inflection at x=0 and a unique real solution for the crossing with any y=mx, m

[@ianfowler9340]

Note that for any cubic, if you find the equation of the tangent at the point of inflection, x = a, then the equation of the intersection for the tangent and the cubic will have a triple root at x = a. Holds true for any polynomial, not just cubics.

[@bjornfeuerbacher5514]

"By expanding sin(x) in a truncated Maclaurin series" But you can only do that if you already know the first (and higher) derivatives at x = 0 ?!

[@pwmiles56]

@@bjornfeuerbacher5514 Quite true! I was just exploring the idea. Like I say the approach would need tightening up. I suspect there is not much to be done with transcendental functions, as they are defined as limits anyway.

[@seitanarchist]

It seems to me that refining this definition would involve asserting uniqueness of the solution in some open interval I around x = a, then perturbing the slope m a little in *both* directions, and requiring that the solution x = a to the equation is *not* unique in the interval I for any sufficiently small perturbation in at least one of the directions. But this now sounds dangerously close to defining the derivative using a limiting process. This would also would only generally work for functions that are continuous in a neighborhood of x = a. The classic pathological example: f(x) = x^2 if x is rational ; f(x) = 0 if x is irrational satisfies f ' (0) = 0, but the equation f(x) = 0 has infinitely many solutions in any neighborhood of x = 0.

[@seitanarchist]

I also just realized there are issues with the function f(x) = x^2 sin (1/x) if x != 0 ; f(0) = 0. Again, f ' (0) = 0, but the equation f(x) = 0 has infinitely many solutions in any neighborhood of 0. This function is not quite as pathological as the last one, either.

[@seitanarchist]

And yeah, to follow up: this definition only works for analytic functions, in general. Even smooth/C^infinity functions can have infinitely many zeroes in a neighborhood of a point, without being identically zero in that neighborhood. Any such function is one whose graph, at some point, does not admit a unique local intersection with its tangent line. And of course, affine functions also do not produce unique intersections. Those can easily be brushed aside in establishing the definition of the derivative, but throwing out all non-analytic functions is harder to justify.

[@mattikemppinen6750]

what's up with the discord? all discussion channels gone since spring or so?

[@emjizone]

10:14 Ok, so it's nothing but taking the infinity concept from the lim function and stuff it into numbers. "Nothing" ? Mmm… maybe not. I see interest in this in type theory and binary implementations.

[@kikilolo6771]

1:56 how do you come up for this tangent formula without defining the limit ? And if you define it by that dformpula how do you prove it has only one intersection ?

[@redhotdogs3193]

When does it work if we define the tangent as the line through the curve s.t. the curve is above or below the line except at the point itself for some neighborhood? (And define d/dx as the slope)

[@soupy5890]

If we're talking a tangent line as being a line that locally only intercepts at the point, one function where that would not exist where the derivitive would be a function that oscillates infinitely close to a point: like f(x) = sin(1/x) times x^2 at x=0

[@methatis3013]

There are functions that have discontinuous derivatives

[@ruferd]

This is a neat method! It seems like something that they would have done historically before Newton and Leibniz (did Fermat or Descart ever try this?)

[@ianfowler9340]

Both. This is Descartes' Method in disguise. Fermat also had a way of finding the slope that involves "almost" similar triangles and substituting 0 back in for "E" after cancelling "E" as a common factor in a ratio. Very similar to Newton. Pretty amazing.

[@goodplacetostop2973]

15:48

[@eiseks3410]

You're a freak

[@arantheo8607]

How should tangent be defined?

[@flavioxy]

isn't that how euler did it?

[@michaelwarnecke3474]

I think I am missing something really fundamental here, but in the case of sqrt(x) at x=1 for example, wouldn't the line 1 - (x-1) also work? It coincides with f in exactly one point, (1,1).

[@Apollorion]

But is that line (i.e. y=1-(x-1)) also a _tangent_ line?

[@michaelwarnecke3474]

@@Apollorion It's not, but it still satisfies the condition, right?

[@Apollorion]

@@michaelwarnecke3474 That means that that condition does not always imply that when the line y=f(a)+c(x-a) intersects with the curve y=f(x) only once, that that line is the tangent line, because your example line y=1-x(-1) is an counter example. N.b. we're looking for that tangent line because via the other definition of the derivative, the slope of this line is the derivative which we're looking for.

[@HappinessReborn]

Can't you define the derivative as being the unique unary operator that takes a function and returns another function (with the usual derivative properties)?

[@HappinessReborn]

Or maybe returns a partial function? (For cases like the modulus as an input)

[@methatis3013]

How is it a unique operator? There are infinitely many unary operators (even linear ones) that have a function as an input and a function as an output

[@HappinessReborn]

@@methatis3013 But if you add the properties of the derivative as an axiom of said function, maybe it becomes unique?

[@methatis3013]

@@HappinessReborn what properties do you mean? Linearity? There are, again, infinitely many functions that satisfy linearity, and they don't even need to be on differentiable functions. A set of all linear operators on functions from R to R is a vector space in itself

[@HappinessReborn]

@@methatis3013 properties like the sum property of derivatives and multiplication property of derivatives. Is there more than 1 function that satisfies both properties?