The Most Overlooked Concept in Calculus - Calculus of Inverse Functions

In this video, we look at one of the most overlooked concept in calculus, which is the derivatives and the integrals of inverse functions.

Chapters:
00:00 Inverse Functions (Intro)
01:54 Finding Inverse is Hard (Intro)
04:37 Derivative of Inverse Functions
06:28 Integral of Inverse Functions
08:38 Using Them to Solve Challenging Problems
09:58 Applications (Outro)

Music🎵:

Corrections:
2:16 omega should be [-1+sqrt(-3)]/2 (*principal cube root of unity)
4:13 denominator should be f'(f inverse of x)

[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

[ArminVollmer]

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

[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.

[hanspeter]

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

[tejavvo]

If you ever make a series on slightly more advanced calculus, ill definitely watch it

[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!

[ingiford]

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.

[Craig]

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

[liamfeatherly]

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.

[danielkeliger]

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.)

[algorithminc.]

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

[tolkienfan]

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

[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! 🥺

[xminty]

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

[netizenkane]

Nice! Really hope you cover the probability transform.

[Calcprof]

Youngs inequality; One of my favorites.

[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 <3

[lperezherrera]

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

[AlbertTheGamer-gksn]

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,

[bishan_]

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