As a computer scientist, after reading a shit ton of physics papers (don't ask me why), I can certainly conclude that physicists have this innate talent of pulling out random equation manipulations out of their asses like it's some voodoo black magic.
Wait until you've seen the tricks mathematicians pull-defining concepts so abstract that they might quite literally exist only on a conceptual level, thats true black magic. This abstraction is one of the reasons so many theories emerge; mathematicians manipulate or alter formulas to explore the outcomes, often leading to seemingly ridiculous results. Yet, these results sometimes find real-world applications beyond mere calculations, complex numbers being a prime example.
@@GabriTell me to my Physics Professor: YOU CANT JUST CANCEL DERIVATIVES WHAT ARE YOU DOING. Him: it gets you the right answer anyways, we dont care about the complicated math here
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).
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)
Just stumbled upon this on my algorithm, and I gotta say I did not expect that this made me watch the whole video. Now I am waiting for more videos like these on the channel
This is so freakin' cool! Not just the anime characters showing us math, but the actual idea of a differentiation technique using square root instead of just x! Thank you!
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.
I am watching this video in Korea. I am a middle school student, so it is a little difficult to understand, but it is fun because it is explained kindly. I will watch it often in the future. Thank you!
A good refresher, and i don't think I've ever derived with respect to the root of a variable. Oftentimes it's easy to forget that d/dx actually has more meaning than derive with respect to x. This a good segway to Partial Derivatives as well. Very crucial in our 3d world. And i do love my skeleton equations.
For anyone who wants a rigorous explanation of the "inversion of derivatives" identity, here's a tentative : Let f a bijection defined on I, an open interval. We use the notation y=f(x) for all x in I. Since f is a bijection we also have x=f-¹(y). Basically we want to show: « dy/dx = 1/(dx/dy) » So let's prove that: « dy/dx dx/dy = 1 » The proper way to write this should be: ~~~ For all x in I such that f'(x)≠0 [let's call this - open - set J, from now]: [df/dx](x) × [df-¹/dy](y) = 1 ~~~ Which is the same as saying: ~~~ For all x in J: f'(x) × (f-¹)'(y) = 1 ~~~ And an even better way to write this is: ~~~ For all x in J : f'(x) × [(f-¹)' o f](x) = 1 ~~~ since y=f(x). Remembering that (f-¹)' = 1/(f' o f-¹), when it exists, we have for all x in J: f'(x) × [(1/(f' o f-¹)) o f](x) = f'(x) × [(1/(f' o f-¹ o f)](x) = f'(x) × [1/f'](x) = f'(x)/f'(x) = 1 Therefore, physicists were always right on that one BUT there are implicit hypothesis which you should keep in mind: ★ f has to be a bijection on the considered set ★ f' cannot be 0 anywhere you want to differentiate f-¹ (Oh and by the way, let's not overlook physicists or more generally non math STEM students. Math is a great thing but it's even better when you learn how to adapt it to your domain.)
I really like the freedom one gets with the leibnitz notation of derivatives. With implicit differentiation and Chain rule, fractal derivatives actually seem really trivial with physicist's notation.
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!)
I have a Masters in Math. And I am always looking for a way to explain concepts that my students can/might follow. Zundamon's Theorem does not disappoint. But I must say, sometimes I feel I am on a magic carpet ride, and the Anime girl on the right says...."It is true." And it is!
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.
@@Yubin_Lee_Doramelin I would like to make a video on that, its just that I dont know how to come up with animations and stuff @zundamon's theorem are you interested in a colab?
6:53 It may seem like it's an abuse of notation, but i assure you it isn't. I remember from Diff. Equ that there are conditions and rationale that allow this. Your friendly Physicist
6:34 you can do that. But it's called *chain rule*. You can prove that using the limit definition of derivative: df/dg = lim[h -> 0] (f(x+h) - f(x)) / (g(x+h)-g(x)) Multiplying by 1... or h/h df/dg = lim[h -> 0] (h(f(x+h) - f(x))) / (h(g(x+h)-g(x))) Rearranging: df/dg = lim[h -> 0] [(f(x+h) - f(x)) / h] * [h / (g(x+h)-g(x))] df/dg = lim[h -> 0] [(f(x+h) - f(x)) / h] / [(g(x+h)-g(x)) / h] Assuming df/dx and dg/dx exist and dg/dx is different from 0... then this limit can be decomposed into the division between 2 limits: df/dg = [lim[h -> 0](f(x+h) - f(x)) / h] / [lim[h -> 0](g(x+h)-g(x)) / h] These two limits are the definition of df/dx and dg/dx... which we assumed exist df/dg = [df/dx] / [dg/dx] The statement for the chain rule is slightly different: df/dx = [df/dg] * [dg/dx] But you can arrive at that by dividing both sides by [dg/dx]
At [8:37], d x^beta / d x^alpha = beta/alpha x^(beta-alpha) Derivating again with respect to x^alpha: d [beta/alpha x^(beta-alpha)] / d x^alpha = beta/alpha * (beta-alpha)/alpha * x^(beta-alpha-alpha) = = beta*(beta-alpha)/(alpha^2) * x^(beta-2*alpha) For the particular case alpha=1/2, the result does not match the expected result, which is beta * x^(beta-1) A better definition for the fractional derivative of x^beta, that does not have this problem, is d x^beta / d x^alfa = Gamma[1+beta]/Gamma[1+beta-alpha] * x^(beta-alpha) Derivating again with respect to x^alpha: d [Gamma[1+beta]/Gamma[1+beta-alpha] * x^(beta-alpha)] / d x^alpha = = Gamma[1+beta]/Gamma[1+beta-alpha] * Gamma[1+beta-alpha]/Gamma[1+beta-alpha-alpha] * x^(beta-alpha-alpha) = = Gamma[1+beta]/Gamma[1+beta-2*alpha] * x^(beta-2*alpha) For the particular case alpha=1/2, the result matches the expected result, which is beta * x^(beta-1)
=D/d√x f((√x)²) =d/d√x f(g(√x)) such that g(x)=x² Use the chain rule =g'(√x)f'(g(√x)) Since √x is our varible it is easy from here g(√x)=√x²=x g'(x)=2x Thus g'(√x)=2√x =2√x f'(x)
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.
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
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]
"Fractal derivatives" and "fractional calculus" are too confusing. They not only contains unorthodox differentiation, but also, in Wikipedia, there's even an extended thing called "fractal-fractional calculus"...
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.
I came up with an argument using the chain rule. From the chain rule, df/dx = df/dg * dg/dx. Let g(x) = x^a, then dg/dx = ax^(a-1) from the product rule. So, df/dx = df/dx^a * dx^a/dx. Rearranging, df/dx^a = (df/dx) / (dx^a/dx). Substituting the dx^a/dx result from before gives: df/dx / (ax^(a-1)) = (1 / ax^(a-1) * df/dx
![Noob To Max With DRAGON REWORK In Blox Fruits [FULL MOVIE]](http://i.ytimg.com/vi/LnBMOinoOvA/mqdefault.jpg)
As a computer scientist, after reading a shit ton of physics papers (don't ask me why), I can certainly conclude that physicists have this innate talent of pulling out random equation manipulations out of their asses like it's some voodoo black magic.
as a game programmer i agree whole heartedly
Wait until you've seen the tricks mathematicians pull-defining concepts so abstract that they might quite literally exist only on a conceptual level, thats true black magic. This abstraction is one of the reasons so many theories emerge; mathematicians manipulate or alter formulas to explore the outcomes, often leading to seemingly ridiculous results. Yet, these results sometimes find real-world applications beyond mere calculations, complex numbers being a prime example.
math really is the art of pulling shit out of your ass
As a CS major I can agree
And somehow they get it right 💀😭
substitution is one hell of a drug
6:44 is literally any non math STEM student. "I see symbols above the bar and below the bar, so it has to be a fraction. Who cares about rigour"
An anime girl teaching me math is something I needed.
fr
UHM ACKSHUALLY ZUNDAMON IS A VOICEBANK
Exactly bro
6:44
Most rigorous physics student.
Physicists: "We don't need mathematicians to advise our papers"
Also physicists when they're left alone:
@@GabriTell me to my Physics Professor: YOU CANT JUST CANCEL DERIVATIVES WHAT ARE YOU DOING. Him: it gets you the right answer anyways, we dont care about the complicated math here
Seeing zundamon explain calculus for me isnt something i knew i needed, thank you.
She did NOT just say "I understand it now" LOL 0:10
IKR?
I love these videos. Combining two things I love, maths and zundamon
Bashame left the rice fields and started math fields
is this where these mysterious fields medals I've been hearing about come from? jiang ping leaving the mines for greener pastures with this one
Deer God she’s actually fully awake
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).
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)
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
learning in class ❌
learning from zundamon ✔
I love the little end screen with a quote!
i love these interesting/unusual problems and the video format, keep up the good work
I love this channel. Keep up with this great work
Just stumbled upon this on my algorithm, and I gotta say I did not expect that this made me watch the whole video. Now I am waiting for more videos like these on the channel
This is so freakin' cool! Not just the anime characters showing us math, but the actual idea of a differentiation technique using square root instead of just x! Thank you!
anime girls/vtubers explaining math. I never knew I needed this in my life
ai anime girls you should say
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.
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 great haha, so much different than the usual math videos
I graduated a few years ago and these videos heal my soul. Thank you!
0:11 "i understand it now" 🗣🔥
I am watching this video in Korea. I am a middle school student, so it is a little difficult to understand, but it is fun because it is explained kindly. I will watch it often in the future. Thank you!
0:12 Zundamon: “Help me, Eirin!”
I really enjoy this format for some reason
A good refresher, and i don't think I've ever derived with respect to the root of a variable.
Oftentimes it's easy to forget that d/dx actually has more meaning than derive with respect to x.
This a good segway to Partial Derivatives as well. Very crucial in our 3d world.
And i do love my skeleton equations.
love u guys and ur work
this zundamon girl is like really clever isn't her? she must be some kind of genius
For anyone who wants a rigorous explanation of the "inversion of derivatives" identity, here's a tentative :
Let f a bijection defined on I, an open interval.
We use the notation y=f(x) for all x in I. Since f is a bijection we also have x=f-¹(y).
Basically we want to show:
« dy/dx = 1/(dx/dy) »
So let's prove that:
« dy/dx dx/dy = 1 »
The proper way to write this should be:
~~~
For all x in I such that f'(x)≠0 [let's call this - open - set J, from now]:
[df/dx](x) × [df-¹/dy](y) = 1
~~~
Which is the same as saying:
~~~
For all x in J: f'(x) × (f-¹)'(y) = 1
~~~
And an even better way to write this is:
~~~
For all x in J : f'(x) × [(f-¹)' o f](x) = 1
~~~
since y=f(x).
Remembering that (f-¹)' = 1/(f' o f-¹), when it exists, we have for all x in J:
f'(x) × [(1/(f' o f-¹)) o f](x)
= f'(x) × [(1/(f' o f-¹ o f)](x)
= f'(x) × [1/f'](x)
= f'(x)/f'(x)
= 1
Therefore, physicists were always right on that one BUT there are implicit hypothesis which you should keep in mind:
★ f has to be a bijection on the considered set
★ f' cannot be 0 anywhere you want to differentiate f-¹
(Oh and by the way, let's not overlook physicists or more generally non math STEM students. Math is a great thing but it's even better when you learn how to adapt it to your domain.)
I am from college and my family watching this would be wild help!!!
It's actually an amazing content
I really like the freedom one gets with the leibnitz notation of derivatives. With implicit differentiation and Chain rule, fractal derivatives actually seem really trivial with physicist's notation.
HOLY SHIT YOU MAKE ENGLISH VERSIONS I LOVE YOU INSTASUB
Damn, I have so many pwoblems alweady and this just added fuel to the fiwe.
I have no idea what happened but I enjoyed the funny characters yapping
Ahh this! If one goes down the rabbit hole and tries to be formal, this could well end at some branch of the modern math
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!)
Why am I watching this at 3am and why can't I stop until it finishes.
I have a Masters in Math. And I am always looking for a way to explain concepts that my students can/might follow. Zundamon's Theorem does not disappoint. But I must say, sometimes I feel I am on a magic carpet ride, and the Anime girl on the right says...."It is true." And it is!
Best math content
What have I ended up on.
Why are people so bewildered from differentiating with a function 😭😭😭😭
Can confirm, works very well
Taught me more than my actual school teacher😅
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.
Your anime is so cute and helpful❤, keep up the good work
now I love math
6:34 - Incoming chain rule.
Important in dynamics
hey are you planning to make videos on fraction derivatives? tthat would be an interesting topic too
Absolutely. There is "fractal-fractional calculus" as well, according to Wikipedia...
@@Yubin_Lee_Doramelin I would like to make a video on that, its just that I dont know how to come up with animations and stuff @zundamon's theorem are you interested in a colab?
it is, if I'm not wrong it involves Gamma Functions and it's super easy
@@redcap5616 true, but it has been very helpful in non-newtonian fluids
my brain is melting
6:53 It may seem like it's an abuse of notation, but i assure you it isn't. I remember from Diff. Equ that there are conditions and rationale that allow this. Your friendly Physicist
6:34 you can do that. But it's called *chain rule*.
You can prove that using the limit definition of derivative:
df/dg = lim[h -> 0] (f(x+h) - f(x)) / (g(x+h)-g(x))
Multiplying by 1... or h/h
df/dg = lim[h -> 0] (h(f(x+h) - f(x))) / (h(g(x+h)-g(x)))
Rearranging:
df/dg = lim[h -> 0] [(f(x+h) - f(x)) / h] * [h / (g(x+h)-g(x))]
df/dg = lim[h -> 0] [(f(x+h) - f(x)) / h] / [(g(x+h)-g(x)) / h]
Assuming df/dx and dg/dx exist and dg/dx is different from 0... then this limit can be decomposed into the division between 2 limits:
df/dg = [lim[h -> 0](f(x+h) - f(x)) / h] / [lim[h -> 0](g(x+h)-g(x)) / h]
These two limits are the definition of df/dx and dg/dx... which we assumed exist
df/dg = [df/dx] / [dg/dx]
The statement for the chain rule is slightly different:
df/dx = [df/dg] * [dg/dx]
But you can arrive at that by dividing both sides by [dg/dx]
The fact that Zundamon just compressed that into a single step is CRIMINAL.
Nah but physicist be livin on a higher ground 💀🙏😭🚬🚬
my favorite is:
let f(x) = 3x+2
find d/d3(f(x))
No way we got weeb math before GTA 6
You’ve successfully published Calculus 1 for kids
I like the other one better and just reading the subtitiles
2:03 this is not the denominator
Subbed so hard.
Now I might be confused.
Fractal differentiating is NOT the same as fractional calculus? ... right?
0:01 2root(x)f'(x) ? Maybe thinking of d/droot(x) as a fraction helps
6:59 uhhh dam 😅
we would have technology equivalent to magic already if the internet is not filled with brainrot
I though of the substituion method too
So it only took an anime girl to make me focus so hard
At [8:37],
d x^beta / d x^alpha = beta/alpha x^(beta-alpha)
Derivating again with respect to x^alpha:
d [beta/alpha x^(beta-alpha)] / d x^alpha = beta/alpha * (beta-alpha)/alpha * x^(beta-alpha-alpha) =
= beta*(beta-alpha)/(alpha^2) * x^(beta-2*alpha)
For the particular case alpha=1/2, the result does not match the expected result, which is beta * x^(beta-1)
A better definition for the fractional derivative of x^beta, that does not have this problem, is
d x^beta / d x^alfa = Gamma[1+beta]/Gamma[1+beta-alpha] * x^(beta-alpha)
Derivating again with respect to x^alpha:
d [Gamma[1+beta]/Gamma[1+beta-alpha] * x^(beta-alpha)] / d x^alpha =
= Gamma[1+beta]/Gamma[1+beta-alpha] * Gamma[1+beta-alpha]/Gamma[1+beta-alpha-alpha] * x^(beta-alpha-alpha) =
= Gamma[1+beta]/Gamma[1+beta-2*alpha] * x^(beta-2*alpha)
For the particular case alpha=1/2, the result matches the expected result, which is beta * x^(beta-1)
Note that Gamma[1+beta]=beta!
Agree or disagree?
=D/d√x f((√x)²)
=d/d√x f(g(√x)) such that g(x)=x²
Use the chain rule
=g'(√x)f'(g(√x))
Since √x is our varible it is easy from here g(√x)=√x²=x
g'(x)=2x
Thus g'(√x)=2√x
=2√x f'(x)
In fact using the chain rule u can generalise this further
d/dg(x) f(x) using the chain rule is the same as (d/dx f(x))/(d/dx g(x)) =f'(x)/g'(x)
I used to do this on my free time though with integrals XD
my little brain :0
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.
5:07, wait, wait, wait, this is just 1/dsqrt(x)/dx
Yes mom, I am studying
Question: Differentiating with respect to... What
Answer: Making love with Zundamon
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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?
damn, ok, I'll just go fuck myself then ._.
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
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_
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]
do you mayhaps, perchance mean fract *ion* al differentiation??
just chain rule
7:00 how did you get the last equation?
hizo la derivada de la potencia, al dar vuelvta dx/dx^a ---> dx^a / dx ==== dy /dx ---> y = x^a ----> D(y) = D(x^a) ====== dy/dx = a * x ^ (a - 1)
JUST DIFFERENCIATE UPPER PART AND LOWER PART IN FRACTION. EZ .
Isn't this just the 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?
Just to point out to other people watching the video
This is not fractional derivatives it's fractal and they are not the same
"Fractal derivatives" and "fractional calculus" are too confusing. They not only contains unorthodox differentiation, but also, in Wikipedia, there's even an extended thing called "fractal-fractional calculus"...
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
one of my classes has me differentiating with respecto to 1/T :(
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
If the channel had less than 1000 subscribers when you commented, it grew real fast
Could this be generalised to d/d g(x) f(x) = (f(x+h) - f(x)) / (g(x + h) - g(x)?
Bro I suck at math, I literally can't do any of this
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。
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
What have i found 💀💀💀
What is the music on the background?
Please check the video description👍
Bruh Math no doing it
UwU voice 😭🤚
Shit
sqrt(x)=t; d/dt of t^2=2t=2sqrt(x). DONE IN 5 SEC
I came up with an argument using the chain rule. From the chain rule,
df/dx = df/dg * dg/dx.
Let g(x) = x^a, then dg/dx = ax^(a-1) from the product rule.
So, df/dx = df/dx^a * dx^a/dx.
Rearranging, df/dx^a = (df/dx) / (dx^a/dx).
Substituting the dx^a/dx result from before gives:
df/dx / (ax^(a-1))
= (1 / ax^(a-1) * df/dx