The Most Overlooked Concept in Calculus | Calculus of Inverse Functions

[@ArminVollmer]

There is a prime tick missing, in the upper formula at 4:19

[@minecrafting_il]

mmm crunchy symbols edit: holy shit the idea of creating an arbitrary probability distribution was something I messed with just a few days ago this is creepy

[@Vsevolodbochkov]

Bro is NOT crunchycat Luna💀

[@barakap2372]

ישראלי?

[@minecrafting_il]

@@barakap2372 כן

[@awildscrub]

This channel is so underated! Having geometric interpretations and visualizations make the concepts so much easier to understand, precisely the way math should be learned. Unfortunately, most teachers don't provide this kind of explaination as most of them don't even understand these interpretations themselves, only with memorization. Keep up the good work!

[@rugbybeef]

Thank you, this was a very clear and concise explanation of these ideas.I hope that high school calculus teachers see this and adopt it in their pedagogy. This view of integration by parts and implicit differentiation were deemphasized when I was going through BC Calculus due to our focus on teaching to the test.

[@hanspeter5118]

urf... 9:05 you got me with that beautiful track, love it

[@Fire_Axus]

your feelings are irrational

[@bopcity5785]

I thought I was hearing stuff at first😆

[@hubutaha3556]

me too haha, what is urf in math world?🤣😇

[@liamfeatherly458]

There will never be a video about the most overlooked subject of a branch of mathematics; for the moment a video is made about it it is now by definition no longer the most overlooked as there will always be a subject without a video.

[@ntwede]

You should make a video about that :)

[@HJ28_398]

Just make a video about every single possible subject, duh 🙄

[@ezrasteinberg2016]

Excellent video!

[@ingiford175]

This concept is core to Spivak's trig and calculus on trig function. Using Inverse functions he bypasses having to prove Lim x goes to 0 of sin(x)/x is 1 that most other calculus books need to prove derivatives of trig functions.

[@abebuckingham8198]

Just looked it up and it's so clean. I love it.

[@ingiford175]

@@abebuckingham8198 It relies heavily on derivatives of inverse functions which in other courses are hardly mentioned. You already had both differentiation and integration, while most other books handles trig and exponential differentiation before you move from differentiation to integration.

[@abebuckingham8198]

@@ingiford175 It's common in complex analysis to define the trig functions using their taylor series so they develop a great deal of theory before introducing them. So there is some precedent for deferring them. I especially enjoyed how they derived the addition formula.

[@xninja2369]

It's just basic highschool stuff lamfo isn't that taught school ?🤣

[@ingiford175]

@@xninja2369 Most of my calc books do not do a derivative of a generic inverse (specific functions, they have many examples (ie arctan, etc)

[@Craig2]

Amazing explanation. I never properly understood this concept before I watched this video. Thank you for sharing.

[@AlbertTheGamer-gk7sn]

0:08 There are 3 types of functions: 1. One-to-one function: These functions are STRICTLY increasing or decreasing throughout its domain and has no first-order critical points besides the endpoints if they exist. The inverse function is defined throughout its entire domain. 2. Piecewise inverse function: These functions are not one-to-one, but an inverse exists for a PORTION of the function. These functions have at least one first-order critical point. 3. Singular function: The term "singular" in mathematics refers to anything that are not invertible. Because of that, a singular function is a function that doesn't have an inverse. These functions exist as annihilation functions, where all values in its domain get "annihilated" to a single value as its range. Because of that, a function is singular if and only if (iff) its derivative is 0 throughout its domain, similar to how a singular matrix has a determinant of 0. Let's say you have this function f(x)=1. This means that for every value of x inserted, the function returns 1 every time. Therefore, this graph takes the form of a horizontal line at y=1. If we say we want to invert f(x), we need to switch the domains and ranges, and since this function annihilates all values of x and returns 1, meaning the domain of the function is all numbers, but the range of the function is 1, if an inverse exists, it will then map 1 for the domain and, uh-oh, we have a problem here. A function operates like a time-distance relation: You can be in the same position at 2 different Planck Times, but you, at the same Planck Time, cannot exist in 2 different locations. If f(x)=1, f^-1(x) will have f(n!=1) not making sense, as f(x) cannot not equal 1. However, f^-1(1) maps out all real numbers, meaning that it wouldn't be a function, as at the same time the function simultaneously exists everywhere. This means there doesn't exist any continuous piece of this function that makes sense; it only makes sense for that one point, and even for that point, it violates the basic rules for functions. Graphically, to invert a function, you need to reflect it over the line f(x)=x, and reflecting a horizontal line gives you a vertical line, which violates the vertical line test. Finally, using singular functions, we are able to introduce a concept called indeterminate forms. As you can see, for the function f(x)=1, f^-1(1) gives an indeterminate form due to it being a vertical line at x=1. Given a singular function f(x), f^-1(f^(x)) is indeterminate. Even though applying the inverse function after applying the function should result in the identity function f(x), for annihilation functions, it all has lots of paradoxes. An indeterminate form comes from basic algebra, where a declared variable has no value assigned to it, as variables are MEANT to be indeterminate until you assign a certain value of it or a transformation of it. For example, x is indeterminate. With n variables, you need n equations to make the values determinate. Some equations are degenerate and continue to evaluate an indeterminate form, such as x=x, as it is true for all x. Using that, we can prove the 11 indeterminate forms are indeterminate using these formulas: 0x = 0 for all Aleph-Null x, so 0 / 0 is indeterminate. 0 * infinity and infinity / infinity are variants of 0 / 0, as 0^-1 = infinity. infinity + x = infinity for all Aleph-Null x, so infinity - infinity is indeterminate. The infinitieth root of x is 1 for all Aleph-Null x, so 1^infinity is indeterminate. 1^x = 1 for all Aleph-Null x, so log_1(1) is indeterminate. 0^x is either 0, 1, or infinity, so log_0(0), log)_0(infinity), log_infinity(0), and log_infinity(infinity) are all indeterminate. Using these equations, we can prove which forms are NOT indeterminate, to see how many indeterminate forms are there. Calculi operate on indeterminate forms, as a derivative is equal to 0 / 0, and integrals are 0*infinity, so we can see how many calculi are there. For example, the product/geometric calculus consists of the product derivative, which is limit as h goes to 1 of log_h(xh/x), which is log_1(1), and the product integral, which is 1^infinity. Infinity is a fixed-point of X = X+1, so when dealing arithmetic with infinity, we use a variable x to denote the fixed-point, which is useful on checking if forms are indeterminate. Case 1: 0*infinity, x=0*infinity=0+0+0+0+0+...+0+0+0+0=0+(0+0+0+0+...+0+0+0+0), so x = x+0, and x = x. Indeterminate. Case 2: 1^infinity, x = 1^infinity = 1*1*1*1*1*...1*1*1*1*1=1*(1*1*1*1*...1*1*1*1*1), so x = 1*x, and x = x. Indeterminate. Case 3: 1^^infinity, x = 1^^infinity=1^1^1^1^1...1^1^1^1^1=1^(1^1^1^1^1...1^1^1^1^1), so x = 1^x. This only is true if x=1, so 1^^infinity is NOT indeterminate. However, if x=infinity, it equals an indeterminate form, so it is considered indeterminate in the complex world. Case 4: 0^infinity, x = 0^infinity = 0*0*0*0*0*...0*0*0*0*0=0*(0*0*0*0*...0*0*0*0*0), so x = 0*x, and x = 0. Not indeterminate in the real world. Case 5: 0^^infinity, x = 0^^infinity = 0^0^0^0^0...0^0^0^0^0=0^(0^0^0^0^0...0^0^0^0^0), so x = 0^x, but NO SOLUTIONS EXIST. This is because 0^^x = 1 if x is even, and 0 if x is odd. Same thing goes with 0^^^x, 0^^^^x, and 0{n}x. In Boolean algebra, if a function has no solutions, it evaluates to false. If it is an indeterminate form, it evaluates to true. If it is only true for a value or a finite set of values, it returns that value or set of values. Therefore, 0/0=0*infinity=infinity/infinity=infinity-infinity=infinity^0=1^infinity=log_1(1)=log_0(0)=log_infinity(0)=log_0(infinity)=log_infinity(infinity)=True

[@Archimedes_Notes]

Albert the gamer.You are the man.i liked this name. This is Archimedes from.Archimedes notes channel

[@AlbertTheGamer-gk7sn]

@@Archimedes_Notes Thank you very much!

[@isakhammer6558]

YES LOOK FORWARD TO NEXT VIDEO

[@danielkeliger5514]

I’m not sure if chi distribution is in the standard libraries, but chi^2 definitely is. Generating Boltzman variables would simply involve taking the root if chi^2 variables. (Intuitively, that is how one derives the Boltzman distirubtion anyways as the components of the velocities are i.i.d. normal.)

[@EpsilonDeltaMain]

Exactly, my next vid is on transformations of rv, which you already know

[@That_One_Guy...]

@@EpsilonDeltaMain correct me if i'm wrong but wouldn't F(f^{-1}(x)) just be equal to x^2/2 ? since f(f^{-1}(x)) = x

[@tolkienfan1972]

Stopped too soon! I want to see how you go from your desired pdf to the prng algo to sample it! Great video

[@iyziejane]

First discretize the domain of the PDF into small bins. Then select a bin uniformly at random. This bin X is associated with a probability P(X) from your PDF. Generate a uniformly random real number A between 0 and 1. If A < P(X), keep the bin you chose in the first part. Otherwise discard the bin X and choose a new one. This is the simplest method and it is called "rejection sampling", and it is an example of a Monte Carlo method. It becomes inefficient if your typical P(X) is very small. When that happens you need a better "prior distribution", and a popular way to do this is with a Markov chain that converges to your PDF using e.g. "Metropolis updates."

[@tolkienfan1972]

@@iyziejane Thanks for the response!

[@xminty77]

very cool video I will be following the rest of your videos ^_^

[@algorithminc.8850]

Thanks. Great video. I look forward to checking out your channel. Subscribed. Cheers ...

[@netizenkane2230]

Nice! Really hope you cover the probability transform.

[@A_doe_wasting_her_life]

omg new upload :DDDdddd

[@scottmiller2591]

4:15 Formula is missing a derivative symbol on f, viz RHS denominator should be f'(f^{-1}(x)), not f(f^{-1}(x)). Otherwise, of course, the function and inverse would simply cancel out.

[@Calcprof]

Youngs inequality; One of my favorites.

[@tomctutor]

Very good. You should first test if the function (which you want to integrate the inverse of) indeed has an inverse (within its stated domain)! How can you do this? e.g 1 Consider f(x) = x^3 - 3x, does it have an inverse function between x = 2 to x = 3? Differentiate, f'(x) = 3x^2 - 3 = 3(x^2 -1), Solve for any f'(x) = 0, stationary points at x= 1, -1, so SPs here are outwith stated domain (also no other critical points) so inverse function exists. Should the domain be chosen so that say, from x=0 to x=2 then the inverse function would not exist. e.g 2 Find def. integral of inverse function of (x^3+x+1) between x=1 to x=3? Before you go ahead and try to do this you should test if indeed the inverse f^(-1)(x) exists? f'(x) = 3x^2 +1 which is irreducible and has no stationary points, so yes it exists everywhere on ℝ. Hope that helps! 🥺

[@eqwerewrqwerqre]

Very nice, though I'm kinda confused on the steps at/before 8:14. Are we just rearranging the above equation to get an expression for the red term? And are we also setting alpha to 0 so the alpha coefficient disappears? It's a very nice presentation, though i Thin using maybe a couple more colors even might help keep things straight for me. Personally i once saw a video that utilized a different color for every term (like 6) and it was a revelation. What would have been very difficult to follow became very easy, i knew exactly what was going where. With so much green and red in equations, in bounds, and as areas in a graph all on the same screen i think my brain is getting kinda lost. That said, i love you for making this video. Please keep making more as long as it makes you happy. I can tell you've put a LOT of work into this and it shows. Thank you so much

[@scottmiller2591]

The definite integral by parts form is intuitive if you draw a rectangle on (a,f(a)), (b,f(b)) - the area of the rectangle is (b-a)(f(b)-f(a)), that is, the first term of integration by parts. The area above the curve in the rectangle is the 2nd term in integration by parts, and the formula is obvious if you flip the axes. The integral we are interested is then obviously the area of the rectangle, minus the area above the curve in the rectangle. The indefinite integral giving rise to this is then obvious as well. There's no need to deal with (0, f(0) at all, and it just clutters up and confuses the intuition.

[@lperezherrera1608]

Just a small tip, when you derive the formula for the derivative of an inverse function, you're using the chain rule so you're implicitly assuming that the inverse is differentiable, which is not immediately obvious

[@Omega-Content]

Yo the pokemon background music 🎶

[@tolkienfan1972]

Slide at 4:20 seems to be incorrect. Denominator should be f'(...), instead of f(...), no? (If it's hard to read, it's f-prime

[@tolkienfan1972]

@user-ky5dy5hl4d it's as we expect later. So I assume it's just a typo

[@Nzargnalphabet]

4:19 I know I haven’t taken calculus yet, but why does the derivative of an inverse seem to just be 1/x? I know that most of the time, f(f^-1(x)) is just x, but I know it’s not that simple, did I miss something obvious?

[@2piee]

Obviously, he made a mistake. The denominator should be f ' (f^(-1)(x))

[@ВикторПоплевко-е2т]

4:48 the derivative of logx is 1/(x*ln10)

[@nathanoher4865]

"log" can refer to the logarithm using any base. It is by convention that calculators use base 10 for "log" due to applications in engineering, which is why "ln" is used for the natural logarithm. In many areas of physics, "log" refers to the natural logarithm. In computer science, "log" often means base 2. It is usually not an issue in practice because the difference between the logarithms is a scale factor. When working out math that someone has published, it should be easy to tell what the base is based on other steps, and when using logarithms in software, it is easy to tell what the base is by testing known values (like 10 or e).

[@ВикторПоплевко-е2т]

@@nathanoher4865 only the derivative of ln(x) is 1/x and if you take the derivative of other logarithms you also need to divide by the ln of the base

[@nathanoher4865]

@@ВикторПоплевко-е2т I know but I'm saying that "log" can be used to mean "log10" or "ln", or any other base in theory when the base is not specified. If someone writes on their paper S = k log W and then below it say exp(S*k) = W then it's obvious that the "log" here is base e. If someone writes on their paper pH = -log [H+] and then below that say [H+] = 10^(-pH) then it's obvious that the "log" here is base 10.

[@ВикторПоплевко-е2т]

@@nathanoher4865 if the base is not specified it's written log based a and it's derivative is 1/(x*ln(a))

[@ВикторПоплевко-е2т]

@@nathanoher4865 My point is that the derivative of the logarithm varies from it's base and if you don't write the base it doesn't make a lot of sense and I can't stop saying it, but log(x) is a logarithm with base 10 and it's derivative isn't (1/x) and you can Google it if you are actually so stupid

[@FadkinsDiet]

4:12 that's a mistake in the derivative of inverse. There should be a ' next to the first f

[@nnsnumbersandnotesunlimite7368]

and what about mentioning the symmetry versus y = x ?

[@ZipplyZane]

4:08 Surely the derivative of the factorial function is defined. We'd need a continuous version of the factorial function to even have a derivative, and thus make that symbol meaningful. And the continuos factorial function is defined using an integral of elementary functions. So surely, by the Fundamental Theorem of Calculus, the derivative would be composed of elementary functions. Where did I go wrong?

[@HaramGuys]

Factorial can be defined analytically as: x! = ∫ t^x e^(-t) dt from 0 to infinity. then d/dx of that would be: (d/dx)x! = ∫ (∂/∂x)(t^x) e^(-t) dt from 0 to infinity. So we can leave it in this integral expression. But if we want to put it in a closed form in terms of elementary functions and factorial function (gamma function with a shift), we are out of luck. In fact, if you try asking wolfram alpha what the derivative of x! is, it will give you something in terms of a new function called digamma function. By definition, digamma function is something that involves the derivative of gamma function in the first place. Notice how the integral is with respect to dt, and x is a parameter of the integrand. Derivative d/dx can "undo" an integral of the form ∫ f(t) dt from 0 to x. but this is not the case.

[@HaramGuys]

Search Leibniz integral rule on wiki and see how to take a derivative of an expression involving integrals. Fundamental theorem is a one specific form of this generalized rule

[@ZipplyZane]

@@HaramGuys So, to put it simply, the derivative of y=x! would be dy/dx, but the integral definition of x! uses dt. And dy/dt ≠ dy/dx

[@bloom945]

Nice!

[@jakeaustria5445]

Thank you. Can I define inverse functions using lambda-calculus?

[@21pandas71]

good job

[@tomkerruish2982]

What about Donald in Mathmagic Land? Seriously, it influenced my eventual decision to study mathematics.

[@drrenwtfrick]

funnily enough the inverse of sin(x)+x when placed in the cosine fucntion leads to a cycloid curve

[@jessewolf7649]

The red equation at 4: 28 appears to have an error. Differentiating the right side yields. x/ [f’(f inv (x))] + f inv (x) - x which does not equal integrand f inv(x) on left hand side of equation.

[@HaramGuys]

chain rule on second item (d/dx)(F(finv(x))) = f(finv(x)) * (d/dx)(finv(x)) = x/f'(finv(x))

[@michaelnash5542]

Wonderful video! Is there any chance you can use colours other than red and green to distinguish areas and algebra? Those of us with the most common type of colour blindness can't see the different (blue and yellow is good, or one solid and one hatched). When you're talking about manipulating the red and green parts of the algebra, I don't know what you mean until you do it and then I'm trying to play catch up with what I missed

[@EpsilonDeltaMain]

Sorry about that. I try to use as many colors as possible in my videos, its pretty easy for me to run out of palettes and have red and green in the same screen. What I could do for sure is to avoid color description that could be confusing for people with color blindness like "red expression" or "green area" and give a more geometric description like "area to the left of the curve"

[@michaelnash5542]

@@EpsilonDeltaMain that is a bit helpful, but doesn't help with the colour coding of algebra. In this video you match ordinates and areas to algebraic expressions by matching red and green colour, and that won't work for many of us I'm afraid EDIT: what might help is to connect parts of diagrams (areas, ordinals) with algebraic expressions by lines, but I don't know if your software can do that easily. Many thanks for your content, it's well made and well explained

[@kasiphia]

I thought a function has to be one-to-one *and* surjective to have an inverse, not just one-to-one?

[@HaramGuys]

Idea of codomain and set theoretic construction of function is far beyond the scope of this video. To an average high school student, range means image

[@IsomerSoma]

As long as the function is injective you can always restrict its range to its image so it becomes bijective. Thus surjectivity is the trivial part while injectivity is the crucial part for invertiblity.

[@IsomerSoma]

​@@HaramGuys you are exaggerating a lot

[@1495978707]

You mixed up onto and one to one. One to one means bijection. Onto means surjective

[@kasiphia]

@@1495978707 Wrong. One-to-one does not mean bijection, it means an injection. Onto means a surjection.

[@spudhead169]

The easiest way to get a standard distribution with an RNG is just to generate two or more and add them together.

[@Nzargnalphabet]

Finally!

[@bishan_8617]

explainin why finding inverses is hard while my brain is just thinking of toothless dancing because of the music lmao

[@Gamr-bc6kp]

soon

[@MAKiTHappen]

Amazing video! I actually have been looking for someone to collaborate with when it comes to explaining mathematical topics like these. Would you perhaps be interested?

[@Thesaddestmomentinourlives]

0:11 isn’t it just multivalued function?

[@tomctutor]

A function cannot map 1:N (one-to-many), it is a mathematical relationship but not a 'function'.

[@Thesaddestmomentinourlives]

@@tomctutor so fucntions like arcsin is just not a function at all? Since it has infinitely many outputs(some angle+2pik)

[@tomctutor]

​@@Thesaddestmomentinourlives No! _arcsin_ returns a single result to any value in its domain, which is [0,1]. In practice the branch to which you choose the output is determined from the CAST diagram [0, 2π) depending on the problem. _arcsin_ , like _arcos_ , can only take a positive input: e.g solve sin(x)= -0.5 arcsin(0.5) = π/6 (Called the 'related angle') so (using CAST) solution is x = 7π/6 or 11π/6. If you want to be pedantic, since angles are modular in 2π you could state; x = 7π/6 ± k2π and so on. Your calculator, or program function, might return garbage if you input any value outside of [0,1] as many of my students have found out the hard way.🤔📐

[@Thesaddestmomentinourlives]

@@tomctutor en.wikipedia.org/wiki/Multivalued_function I meant Arcsin(x), this notation is considered to be multivalued

[@tomctutor]

@@Thesaddestmomentinourlives If we define a function by its descriptor, usually a graph, map or formula to also include the domain. Then I assert for a function to have an inverse (like sin(x) v. arcsin(x)) then it must be 1:1. Restricting the domain to make it so as I described for trig functions. Different in Complex Analysis.

[@megaing1322]

At 2:52, you explicitly said "x to a rational power", but then showed x^x^log(x), which isn't exactly a rational composition? I assume you meant real power? Or am I missing something?

[@hhhhhh0175]

any a(x)^b(x) is elementary under the video's definition because a(x)^b(x) = exp(log(a(x))*b(x))

[@IamJacksHeartCA]

It’s fun to say urf

[@ZipplyZane]

Is there a term for a function that does not include an integral in its definition? In other words, you have an integral where there is no simpler way to express the term?

[@gffhvfhjvf4959]

Lagrange put above Fourier?

[@DarinBrownSJDCMath]

Yeah,... no comparison. Fourier wins hands down.

[@twoonesixsixonetwo]

urf vs erf who win

[@russchadwell]

I used to watch Urf.

[@BLVGamingY]

so if there's an ode that's like f''=-1/(2f) then it's related to urf

[@Fire_Axus]

what do you mean the inverse of e^x is a function?

[@APaleDot]

A function is a operation that maps each input to a unique output. If you look at a function like sin(x), each input has only one possible output. However since sin(x) is periodic, each output is associated with many inputs. Therefore, taking the inverse of sin(x) (switching the input with the output) does not result in a function. You have to limit the input domain to a single period in order to take the inverse. e^x requires no such limiting. The inverse of e^x is ln(x) with no modifications.

[@pauselab5569]

3:12 you mean X^5-X-1

[@pauselab5569]

the way I see it, it's totally fine to not teach quotient and inverse method for integrals because they boil down to special cases of product and u sub.

[@krishivwahi3603]

Literally just got a question where the integral asked was of the form f(x) + f^-1(x)

[@tomctutor]

= ∫ f(x)dx + ∫ f ⁻¹(x)dx

[@BunnyKhatri-pd8zm]

I don't understand shit but shit looks cool maybe one day I will understand of this

[@Grassmpl]

You said *most* functions are not one to one. How do you prove that? Why is there no surjection from family of one to one functions; to the family of all functions?

[@EpsilonDeltaMain]

"Most" as in most functions you have encountered. That is a heuristic statement targeted at high school level, like take any degree 2+ polynomial, it probably has ups and downs. Wasn't thinking about it in terms of cardinality or probability measure or anything like that

[@homomorphichomosexual]

regardless of how epsilondelta meant it, this is actually a really interesting question. there's some stackoverflow discussion ("What is the probability that a random function from N→N is surjective?"), but there seems to be relatively little written about it :( if you find more information on it, please comment in this thread!

[@warguy6474]

​@EpsilonDeltaMain Is it usually via cardinality functions proven as one to one? There was a counting unit in my discrete math course and my professor keeps referencing bijections between counted sets/combinatorics counting and I'm assuming its not quick to explain which is why he didn't explain further on the bijectivity. Similar question goes for one to one functions regarding encoding/decoding data into/from strings

[@ingiford175]

There are many functions where f(a) = f(b) take f(x) = x^2, the values of 2 and -2 both get mapped to 4. Most functions we use that are one to one are monotonically increasing or monotonically decreasing.

[@hhhhhh0175]

most continuous functions aren't one to one, and since this is a video about elementary calculus, we can probably assume almost-everywhere continuity for our functions

[@Arthur-so2cd]

wdym e^x is hiding in the derivative of lnx

[@anjanavabiswas8835]

1/(e^ln(x))) = 1/x. I think.

[@Nzargnalphabet]

Oh, you missed the apostrophe, got it

[@kcrooks7]

You can rotate it 90 degrees

[@nicolasvincent4430]

erf urf

[@user-mf7li2eb1o]

Holy shit, you do better work than 3b1b. You just talk to fast for me….

[@user-mf7li2eb1o]

9:05 WAS THAT A LOL REFERENCE IN MATH???? +6000 Aura

[@MyOneFiftiethOfADollar]

Rather arrogant of you to conclude this is the most overlooked concept in Calculus. The result is an immediate consequence of d/dx(f(f^-1(x)) = x) and the chain rule. I don't know of a colleague who did not use this result to find the derivatives of the inverse trig functions. Oh well, its the norm to exaggerate/lie to get page views.

[@hoteny]

is that just the same as f(y)dy

[@tomctutor]

What is 'that' you are referring to?

[@invisibules]

That music really isn't helping me to concentrate 😯

[@agautam5208]

Go and watch Herbert gross lecture of mit

[@ucngominh3354]

hi

[@04-jayeshkumargupta8]

sneaky jojo refrence

[@algorythm5552]

Stop copying, I had found the demo of the derivative of log(x) before you

[@Archimedes_Notes]

You have an issue representing the area of the inverse function. Therefore ,your geometric interpretation is false. You are very close

[@AlessandroZir]

what's most difficult to get is how much you are just showing off or really intending to convey a useful insight...