these videos have genuinely been such a pleasure to watch, even though I don't know who the anime people are, or why they exist. this is a joy. thank you for making these!
@@fgvcosmic6752Its also that rule which i dont know if it has a name which relates the deravitive of a function to the deravtive of its iverse function. Just because of the use of the leibniz notation the inverse doesnt have to be a function. But its the same logic. The rule on functions states f'(x) = 1/(f^-1(y))'. Notice that f^-1(y) is just df^-1(y)/dy = dx/dy. Thus dy/dx = 1/(dx/dy)
4:20 that is correct 7:00 also correct (even if it a slight abuse of notation) 10:10 By using symbolic expressions by keeping track of the powers of 1/0, we can say that the definition as 1/αx^{α-1}*df/dx is valid (this is basically just analytic functions at x=0) We can also extend this derivative to a derivative with respect to any differentiable function as df/dg=(df/dx)/(dg/dx), however when we have multivariable calculus, we have to sum derivatives with respect to the other base variables, for example derivative of f(x,y,z) with respect to g(x,y,z) can be defined as df/dg=(df/dx)/(dg/dx)+(df/dy)/(dg/dy)+(df/dz)/(dg/dz).
An interesting exercise would be to analyze the process of finding the formula at 7:30 and finding at which step of the reasoning did we assume something restrictive that would not make it work for certain functions like the square root function at 0 for example.
This is how I would approach d/( d g(x) ) f(x). Substitute u for g(x). Rewrite f(x) in terms of u. Differentiate. If the substitution is bijective over some interval, we can write this in terms of x by applying the inverse, g^-1, of the substitution.
4:20 absolutely correct, but the reasoning in a rigorous way is due to topology and im not gonna try to write it in a comment as im not sure yt supports math lingo, and its too long, and its already written somewhere online (lazy too find a source, but i know it exists, written an assay on it a couple of years ago in collage)
also 7:00 also correct, the ability to do so is a very important theorem for differential equations, without it most DEs would be "unsolvable" (in an ease way at least)
also 10:00 is absolutely important! in general if you get an answer, you should still check if it still applies to the original question! (if you let yourself use everything you know without worrying if classical logic is "2way or 1way" then you can quickly get an answer to a question, but the answer can be false so you need to check, as long as you train yourself well, you can insure that you don't "miss" any potential answer and then you are free to blaze trough mountains of exercises! get a tutor if you can and in need of one for this!)
people are always "the derivative isn't a fraction" but even if it isn't a fraction it's a limit of a fraction and lim(a) * lim(b) = lim(a*b) so treating it like a fraction usually ain't even that bad
this channel is cool, too bad it has less than 1000 or so subscribers I believe a channel requires 1000+ subscribers before getting paid, hopefully this gets more views and subscribers. Good math.
Calc BC student here: What is the purpose of taking a fractal derivative? Also, please make a video on partial derivatives. I am interested in learning slightly ahead
Not a mathematician but sometimes its useful when the variable you are differentiating with respect to isn’t a linear function. For example, if you want to find the rate of change of a car moving with respect to time, you would just differentiate it. But if time wasn’t constantly moving at the same speed, but at the rate of square root of x, you could use this formula. Time doesn’t work like that, but it can in some cases. Theres a theorem out there that if 2 particles travel the same path in different time intervals, there exist a function that you can apply on the time variable to make it follow the same “speed” along the path. So if you have 2 cars, and the time function is the square root, you can differentiate car 1 with respect to the square root of x to get the speed of car 2.
Yep, you can just do it by substitution Instead of writing dsqrt(x), we can instead use u=sqrt(x) and write du instead [noting that du = 1/2sqrtx dx = 1/2u du]
substitution is one hell of a drug
6:44
Most rigorous physics student.
these videos have genuinely been such a pleasure to watch, even though I don't know who the anime people are, or why they exist. this is a joy. thank you for making these!
6:34 this looks so illegal, there must have been a physicist involved😅
I didn't do it I swear
I mean, its technically just chain rule and reciprocal rule, so all is well
Just dont try and do that with partial derivatives :D
@@fgvcosmic6752Its also that rule which i dont know if it has a name which relates the deravitive of a function to the deravtive of its iverse function. Just because of the use of the leibniz notation the inverse doesnt have to be a function. But its the same logic. The rule on functions states f'(x) = 1/(f^-1(y))'. Notice that f^-1(y) is just df^-1(y)/dy = dx/dy. Thus dy/dx = 1/(dx/dy)
4:20 that is correct
7:00 also correct (even if it a slight abuse of notation)
10:10 By using symbolic expressions by keeping track of the powers of 1/0, we can say that the definition as 1/αx^{α-1}*df/dx is valid (this is basically just analytic functions at x=0)
We can also extend this derivative to a derivative with respect to any differentiable function as df/dg=(df/dx)/(dg/dx), however when we have multivariable calculus, we have to sum derivatives with respect to the other base variables, for example derivative of f(x,y,z) with respect to g(x,y,z) can be defined as df/dg=(df/dx)/(dg/dx)+(df/dy)/(dg/dy)+(df/dz)/(dg/dz).
An anime girl teaching me math is something I needed.
I love these videos. Combining two things I love, maths and zundamon
Bashame left the rice fields and started math fields
my teacher just said multiply and divide by dx and then replace the dsqrt(x) with dx and calculate the other with the reciprocal differential rule
Not wrong but this gives an intuitive explanation
@@QwertierMannier-yp2hb yeah its just a slight abuse of notation
I love the little end screen with a quote!
An interesting exercise would be to analyze the process of finding the formula at 7:30 and finding at which step of the reasoning did we assume something restrictive that would not make it work for certain functions like the square root function at 0 for example.
love u guys and ur work
You people are cooking just in the way really good videos about understanding math form a student perspective and alwaying people to follow it
This is how I would approach d/( d g(x) ) f(x). Substitute u for g(x). Rewrite f(x) in terms of u. Differentiate. If the substitution is bijective over some interval, we can write this in terms of x by applying the inverse, g^-1, of the substitution.
this is great haha, so much different than the usual math videos
I am from college and my family watching this would be wild help!!!
I love this channel. Keep up with this great work
It's actually an amazing content
I though of the substituion method too
I like the other one better and just reading the subtitiles
Taught me more than my actual school teacher😅
HOLY SHIT YOU MAKE ENGLISH VERSIONS I LOVE YOU INSTASUB
Just to point out to other people watching the video
This is not fractional derivatives it's fractal and they are not the same
my brain is melting
4:20 absolutely correct, but the reasoning in a rigorous way is due to topology and im not gonna try to write it in a comment as im not sure yt supports math lingo, and its too long, and its already written somewhere online (lazy too find a source, but i know it exists, written an assay on it a couple of years ago in collage)
also 7:00 also correct, the ability to do so is a very important theorem for differential equations, without it most DEs would be "unsolvable" (in an ease way at least)
also 10:00 is absolutely important! in general if you get an answer, you should still check if it still applies to the original question! (if you let yourself use everything you know without worrying if classical logic is "2way or 1way" then you can quickly get an answer to a question, but the answer can be false so you need to check, as long as you train yourself well, you can insure that you don't "miss" any potential answer and then you are free to blaze trough mountains of exercises! get a tutor if you can and in need of one for this!)
people are always "the derivative isn't a fraction" but even if it isn't a fraction it's a limit of a fraction and lim(a) * lim(b) = lim(a*b) so treating it like a fraction usually ain't even that bad
this channel is cool, too bad it has less than 1000 or so subscribers I believe a channel requires 1000+ subscribers before getting paid, hopefully this gets more views and subscribers. Good math.
This is an english language version of www.youtube.com/@zunda-theorem
Could this be generalised to d/d g(x) f(x) = (f(x+h) - f(x)) / (g(x + h) - g(x)?
Now I might be confused.
Fractal differentiating is NOT the same as fractional calculus? ... right?
my little brain :0
Isn't this just the chain rule?
市場規模考えたら、もっと英語コンテンツ有っても良いと思うが
日本は世界でもかなり特異な教育体制なので、数学ができる者の率が高いので
有象無象も突っ込んでくるが
(ヨビノリみたいに「私の動画は概ね偏差値60以上を対象としています」と足切りするとか)
海外だと高等教育を受けていないと、数学は無理なので
視聴者もハイエンド寄りになるので、荒れないで済むと思う
Are you calling the rest of the world stupid?
Calc BC student here: What is the purpose of taking a fractal derivative? Also, please make a video on partial derivatives. I am interested in learning slightly ahead
Not a mathematician but sometimes its useful when the variable you are differentiating with respect to isn’t a linear function. For example, if you want to find the rate of change of a car moving with respect to time, you would just differentiate it. But if time wasn’t constantly moving at the same speed, but at the rate of square root of x, you could use this formula.
Time doesn’t work like that, but it can in some cases. Theres a theorem out there that if 2 particles travel the same path in different time intervals, there exist a function that you can apply on the time variable to make it follow the same “speed” along the path.
So if you have 2 cars, and the time function is the square root, you can differentiate car 1 with respect to the square root of x to get the speed of car 2.
Can we substitute f(x) = f((√x)²) lol don't do this, I'm just asking 😊
Yeah , let √x = y , so x = y²
And now it becomes -
dy²
_____ = 2y = 2√x
dy
Yes, but make sure to remember the chain rule!
@@fgvcosmic6752 yeah , half of the _diff._ is just _chain rule_
7:55 noo you cant just split the limit 😭If both are zero or infinity you must use rigerous methods to find the limit
maybe use dual numbers then?
d^m/dx^m(x^n)=(n!/(n-m)!)x^(n-m), n>=m;
d^m/dx^m(x^n)=(Gamma(n+1)/Gamma(n-m+1))*x^(n-m);
If m=1/2 and n=1, the result is 2√x/√π
So when alpha equals -1 that would mean we're integrating it right
d/d(x^-1) isnt integration, it is differentiation with respect to 1/x.
For example, d/d(x^-1) of 1/x = 1, but the integral of 1/x is _not_ 1
Hold up....this is possible even with integration? I'm quite curious
Riemann-Stieltjes Integral, if I'm not mistaken.
Yep, you can just do it by substitution
Instead of writing dsqrt(x), we can instead use u=sqrt(x) and write du instead [noting that du = 1/2sqrtx dx = 1/2u du]
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もしかして日本語版未公開動画ですか？
はい！日本語版も近いうちに公開する予定です。
What is the music on the background?
Please check the video description👍
i gave the respect to the √x 🫡
I am from college and my family watching this would be wild help!!!