delta y vs. dy (differential)
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- Опубликовано: 21 авг 2024
- We will discuss the difference between delta y and dy. Delta y is the difference between y-value value and the original y-value. But dy is the differential and it's an approximated difference between the y-values.
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@blackpenredpen
For those not aware, what he described is essentially Linear Approximation, or Linearization
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I'm pretty sure linearization is different
@@rangertato how so?
@@richardaversa7128 linearization is when you change one or both variables so that the resulting graph is linear, for example a y=1/x graph can be plotted as y vs 1/x instead of y vs x so that it can be analysed without calculus
@@richardaversa7128linear approximation is approximation with tangent line
linearization is modifying curved data to make it fit a linear line of best fit
> ∆y is exact change
that would explain why I never have any ∆y when I go to the coffee shop
genius pun right here, give this person a medal
Under rated
???
Give an explanation please
@@solarsystem1958 del y = change
LMAO
I like these in depth videos, they make you appreciate math more, as apposed to how you learn in high school just rushing through memorizing formulaes and rote learning solutions. Great video!
Yay!! I am glad to hear this. Thank you.
Agreed, I hope he does it more and more
I am happy that we have teacher who at least explained this.
sorry to be so offtopic but does any of you know a way to log back into an Instagram account..?
I was dumb lost my account password. I would love any tips you can offer me!
@Ayaan Landyn Instablaster =)
and now I finally understand why there's a 'dx' tacked on the end of the dif'd function :D thanks!
got my mind blown recently on that part. why did nobody tell me why it is there before?
@@fruityliciousk2704 Because it is Big american secret!
I finally found out what exactly differential is. I've searched on the internet, read books for a few weeks but I couldn't find a tutorial as simple as yours. I'm gonna pray for you. Thank you so much.
Holy moly this was so much better than my professors explanation! He literally was trying to explain the concept using marble and bread.
Interesting, I always understood dy/dx only existed on the limit as Δx approached zero. Anything else was just delta Δy/Δx. ie dy/dx=lim Δx→0 Δy/Δx.
You were correct 100%. (d/dx)() is a function operator that takes a function f as its argument. f has to be a function on x. So (d/dx)(f) is the transformed function. The transformation is defined by the limit of the difference operator, just as you stated. (d/dx)(f) is the same as f'. It's just a different notation.
This is ALL that makes the definition of d/dx, nothing else.
There is NO multiplication by dx or dy whatsoever because d/dx() is ONE symbol. It's like if a teacher would tell you to divide both sides of the equation sin(2x)=1 by s to get in(2x)=1/s.
So, what is done is symbolical manipulations that give you the name of a desired function, e.g. a solution of a differential equation. Hence it's a useful method but the method is formally not mathematical. So, those steps are no more mathematical than, let's say, you make your teacher deliver you a solution of cos(x-sin(×)+sqrt(x))=2 by threatening him to burn his dog. It's a valid and effortless method to get the solution. Once you have it, you can valit it. That threath is a non formally mathematical method just like multiplying by dx.
Importantly, we need to consider why the method works. That reason is of course mathematical. In abstract Algebra operations on objects called dx are defined and it is useful. So, there differentials are from a different perspective.
yes, this is the best way to explain it.
I disagree slightly with the reply above.
While (d/dx)(f) is the 'primitive' notion, its existence implies that of the differentials, and the relationship between the two is more than 'symbolical manipulations'.
If y = f(x), then the differentials of x and y, 'dx' and 'dy', satisfy the equation dy = dx · f'(x) -- this makes it permissible to swap f'(x) with dy/dx (noting that dx is non-zero). It is customary for dx = Δx.
The differentials are then valid objects in their own right, and f being differentiable implies their existence always -- not just 'on the limit'.
The important property at the limit of dx → 0 is that Δy and dy coincide; but the value dy/dx can be considered for non-limit values of dx -- this will in general be distinct from Δy/Δx.
This can be a great introduction into Euler's method as well!
yup!
Mate, thank you a lot. I've been reading James Stewart's book for over an hour trying to understand what you've simplified in 10 minutes.
Been reading Thomas Calculus on it & yea same problem.
It was what I needed today.
This makes my concept clear now! Thanks
Yay!!
Thank you so much!
I'm rewatching this video after 6 months. The idea is very clear, and the explanations were understandable. This helps me a lot.
♥
ruclips.net/video/XQIbn27dOjE/видео.html 💐
Haven't took an intertest in Math ever, even when I was doing this in school.
I always came across the Delta symbol in Wiki articles whenever I was looking up information. This defined it quite well.
With the god pen switching technique he also writes so fast that I didn’t see him writing 7:04 “just an approximated”
A man with so much talent !
This channel is fantastic, I thought that dx and dy were delta x and delta y when delta x approaches zero. That they really make sense only when they are divided by each other and that we can sometimes use them separately but that was just a math trick. Apparently, dx and dy are not necessarily going to zero, they are just delta x and delta y of the tangent line which is determinate by the derivative. The only thing is that ratio dy/dx has to be equal to the ratio of delta y/delta x in the limit when delta x - >0.
I’m stuck because I don’t have a calculator with me 😆
I watched your video to learn actually but I suddenly smile when you're looking and smiling at the camera. How cute of you💕
I love how you show real uses of theoretical calculus. Liked 😉
Thanks. My math teacher just wouldnt tell me what the heck was a dx or dy or d in general (only told me it was differential)
This was a good explaining
That cleared out some of my back of the mind doubts
My boi is slaying the professors with a 10$ mic.
so Δy/Δx gives the exact slope of a secant line, but only the approximate slope of the tangent line. and dy/dx gives the exact slope of the tangent line, but only the approximate slope of a secant line. maybe I'm wrong but this feels like a succinct way to understand the utilities of Δy and dy and the symmetry between them.
Damn. Good thinking
That's a really good way to connect the secant line vs the tangent line.
dy is also a change like ∆y but this change is estimated value. Calculus can pull out very detailed information.
I like the graphs you include to assist your instruction and listener understanding.
Impressed with the way you explained such a technical concept with sweet and smiley face
Thanks
Thank you very much! I am revisiting this topic in Calc 3 and now I get it
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Im gonna cry right now bc i got a 1.5/10 on my calc test bcz of this thing and you just made my life a lil bit better i cant thank you enough fr
Hang in there. Most of us did the same. Just keep doing limit problems...at some point, and it's different for everybody, your brain will thoroughly grasp the concept and application.
Thanks mate by far the easiest explanation vid I have found on RUclips. You must be a great tutor 👍
So, to summarize:
"Delta" y = dy-h
Where h is the difference between two points of a graph/ equation.
As h approaches 0, then "deta y" = dy getting + an ever decreasing value [if the limit exists. ]
I am surprised they don't "totally over shoot" like by 10 or more, then reduce to 5, then 2, then 1 to show the actual process of taking a limit. That way you can see it better...
d(f (x)/dx is a shortway hand of writing delta-y/delta-x when back then they used to find derivatives manually by plugging in delta-x minus h, i recall, they dont have shortcut formula for doing it, now d(f(x)/dy aka dy/dx is now used to implicitly tell that what its doing is approximation
Nice content man, it really helped
Your videos really remind me of just how much I've forgotten. :)
this is linearization
Which is the first two terms of the Taylor Series.
its very useful for me ,ive just confused by the class course ,when i saw this video ,it makes my brain clear again
This is the best explanation ever! Thank you!
His geometric representation is wrong 😊 go and check mit Herbert gross lecture approximation
This is so dope
Thanks so much for these videos
I can't understand math the way is teach it in school
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Thank you Sir....
A student from India...
They skip this in class and say to memorize the formuals...
Do one with Sigma and Integral too!!
You’re a good teacher. Thanks
Can you please share what course are you studying. Your knowledge of calculus is soo coool
WOW! This is one of the best videos on RUclips.
ruclips.net/video/XQIbn27dOjE/видео.html 💐
thank you, you make it so easy to understand what exactly does "d"y mean.
I always thought dx≅0, dy=some function(dx)≅0, but ratio dy/dx gives some real values .... but why we can put dx=1? isn't it was allowed only to put something small like dx=0.00001 instead ?
Then dy will get more accurate.
"small" and "big" are all imprecise and relative notions when numbers are concerned. the closer dx gets to zero, the closer dy gets to Δy (more precise is the approximation). you can arguably put dx equal to any number (1, 0.1, 0.0000...001), and you'll use the result that suits you the most
Thank you for this explanation. It really make me understand better
I was with you until 8:03, when you mentioned that the derivative of the square root of x is one over 2 multiplied by the square root of x.
I know that the square root of x over x will give you the same ratio as the number one divided by the square root of x; but I'm not sure how you got to the derivative you mentioned above.
All late you probably figured it out by now lol but you can write sqrt of x as (x)^1/2. That’s what it is in exponential form. 1 on top power and 2 on bottom means root.
So: (x)^1/2
dy/dx = 1/2(x)^-1/2 bc we move exponent down and subtract by 1.
So rewrite in sqrt form
dy/dx = 1/(2sqrtx)
find dy/dx for y^-3/5=sec^-1*sqrtx* 4^lnx / y^2*lnx* tan^-1 *2x i try solving this derivatives .tanks allot
Δy = f'(x0)Δx + o(Δx) (o(Δx) is little-o of Δx as x -> x0, Δx -> 0)
dx defined as Δx
dy defined as f'(x0)Δx = f'(x0)dx
there is where all difference come from
It's like the difference between summation and integraiton
Thank you sir for posting such great video .I couldn't understand what is the difference between delta and differential.But now iam Clear with that.Such a great explanation ❤
This is from 4 years ago.. How did I never know this?!
This video save my life
And you're naturally funny.
ANOTHER ASIAN MATHEMATICAL GENIUS
so dy/dx is used if you want to find the straight line right? Like, "if the line doesn't bend because of the √x, where would it be if the starting point is at x=4"
Great explanation! Great video.
we're now learning integrals, and I have just found out about this today lol
You explained the difference between delta y and dy. But what is the difference between delta x and dx?
dx is simply lim ∆x when ∆x -> 0
finally i've understood the difference betwen derivatives and differential thaaaaaaanks
Wow, nunca pensé que le entendería a lo que explicaba este man en sus videos, pero me ha servido para la tarea de cálculo xd
Excellent explanation! Thank you!
You are out off this world! Amazing understading you have. I love your videos and the way you just explains stuff my teachers never was capable of.
Thanks very much.
Trying to learn thank you!!
Very concise, very understandable. Thanks.
That always drove me nuts. I finally decided it was change in y over change in x. But instantanious, not when delta x is greater than 0.
wel, as you can see dy is delta y of the tangent line and dx is delta x of the tangent line. Now, that tangent line is determinate by the derivative dy/dx has to be equal to the delta y of a function divided by delta x of the function in the limit when delta x - > 0.
As far as I understand, delta x when x approaches 0 doesn't have meaning on its own, but the ratio has. We are asking what number is the ratio approaching to as the delta x of the function approaches 0. And that "approaching" (which is a jargon word, the real word is limit) is defined by epsilon, delta definition.
@@smrtfasizmu6161
Yes, that is exactly what frustated me, because I to figure something similar out on my own.
@@frankharr9466 I mean that's how yo learn math by figuring stuff on your own. If you watched this video and changed opiniomabout dy and dx you again figured the stuff on your own, hopefully this time with more success. People help you learn by pointing out way your interpretation is wrong and why the actual interpretation is correct. Not having figured it out on your own means you just learned something by heart without understanding. If you understand something you always have a mental picture of it that's only yours and that others can't see. After watching or listening to explanation of other people you can change that mental picture but you can never "download" someone else's mental picture. I remember explaining a friend trigonometry. He was in the same class as I was, he listened to the same professor that I was, but he got different mental image about what is going on then me. Until I figured out exactly what his mental image was, said to him that it was wrong and explained to him that it was wrong, I couldn't help him, he just wouldn't understand. When I was presenting him what's actually going on, his brain tricked him that his mental image is the same as what he was listening. I had to understand exactly how he figured it out and say that it is wrong and why it is wrong. That's not to say that I didn't figure it out on my own just as he did, except that I figured it out correctly. We were both listening to the same stuff in the class and we got different mental images. This is what's happening most of the time when people don't understand a topic, the mental image in their brain doesn't correspond to reality, they don't understand why and how, but they see that their mental picture doesn't make good predictions, so they know it is wrong, they just don't know why. But you can't learn anything without creating mental image in your mind. Understanding something means that your mental image corresponds to reality.
Bravo!! Crystal clear explanation!!
thank you very much. now I actually understand what differentiation tells us.
So, a great oversimplification of this explanation is that dy/dx are the limit where DELTA x -> 0 of DELTA y / DELTA x ?
So underrated, it hurts
Thank you sir for your clear explanation
Exacly saying, for all x and y and a real valued two times differentiable function on [x,y], there exist a number t in (0,1) such that f(x)-f(y)-df(y)(x-y)=1/2 d^2f(y+t(x-y))(x-y)^2. So... the number of correct digits in the approximation is,roughly, the square of the distance |x-y|. Then, if |x-y|
Loved the explanation, really helped me
ruclips.net/video/XQIbn27dOjE/видео.html 💐
I was wondering if, when deriving equations such as momentum equation in aerodynamics, delta P can also be rewritten as dP since both essentially mean change in pressure?
It's amazing your explanation.
Thanks
This video saved my life two times
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Sr Gudiño: Gracias por la observación. Realmente soy nuevo en esta tecnología e ignoro muchos de sus procedimientos. Si lo ofendí de alguna manera, me disculpo. Sin embargo, espero que blackpenredpen me conteste. Estoy muy interesado en los temas que él maneja y sabiendo que es una persona muy inteligente quizá sepa español y pueda enviarme algún mensaje. De cualquier modo, le agradezco su atención.
Sr. Gudiño: Atendiendo a su recomendación, abrí la página Math Stack y también está muy interesante y en ella se puede interactuar, sin embargo, no puedo escribir símbolos, el mismo problema por el cual solicité a blackpenredpen me diera alguna dirección donde pudiera subir archivos. Quizá la solución a mi problema de escribir símbolos en esta plataforma sea muy simple, pero no la se! Acepto sugerencias. Un saludo desde Sinaloa, Mx
I really appreciate your hardwork. Thank you.
ruclips.net/video/XQIbn27dOjE/видео.html 💐
The change in Δy in the second graph is precisely 0.236067977
Very nice explanation!
It's very very clear now. Thank you soooooooo much
ruclips.net/video/XQIbn27dOjE/видео.html 💐
U explaination very well u were very cool throughout the video
Well, the Good Thing is "There is always a Like Button".
Great explanation!
Excellent presentation. vow !!
兄弟，我看了你好多视频了…要是能有中文版本的就跟帅气了～
有中文视频，不过很少，在他频道里找找吧
Excellent explanations.... 👌👌
This type of videos i want 😀
I understand how you defined dy as the approximate change in y-values of the function, but can you say it is the exact change in y-value for the tangent line at the point dy/dx was calculated on y = root(x)?
The way he defined it, yes. It's exactly like Delta y but for the tangent line function instead of the sqrt() function. In any calculus class you will ever take, no. dx and dy are not variables so we can't plug in numbers for dx and see what we get for dy. We can't even find values for dy and dx since they aren't numbers. They're a notation shortcut for certain equations involving limits.
thanks, you are my math god
explaining nicely
Thanks for making the difference clear! :)
OOOOOOHHHHHH WOW.
SO ∆y IS THE GRADIENT FOR THE TANGENT AND dy IS THE GRADIENT OF TWO POINTS RIGHT???
Kind of curious why instead of =DX mathematicians don't use the wavy equal sign Delta X. Doesn't the wavy equal sign mean almost equals to?
I'm somewhat disappointed by the fact that not a single comment about his T-shirt is there in the comment section.
Obrigado pela brilhante explicação.
how he has to laugh, pretending not to know the root of 5 :D love it
very simple to understand.. thanks dear... but the question i have is.. if taking differentiation of a funtion gives rate of change. and called derivative.. and the reverse is called integration..right?? then it has to be the main function again. why it gives the area under curve??. and in this case the origional function in first place suppose to be the area under curve.
excellent explanation, excellent
Great video dude!
what you did to know that the derivative of the square root of x is 1 over 2*sqrx?
danibaracho √x = x^(1/2), so you just use the power rule to get (1/2)* x^(-1/2) which is equal to 1/( 2*√x)