@@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
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!
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!
@Conor Zayden Thanks so much for your reply. I got to the site thru google and I'm trying it out now. Takes a while so I will get back to you later with my results.
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.
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.
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.
@@Davith_D spot on! In simpler terms as ∆x approaches 0, the values of dy and ∆y will appear to coincide, and this takes place at an infinitesimal level, where ∆y is the change in the value of function 'f(x)' and dy is the change in the value of the Tangent line/ derivative function 'L(x)'.This can be understood using the concept of local linearity. If you zoom closely into a function 'f(x)' which is a differentiable curve, the function tends to look and behave like a line around a certain point on the function 'f(a)'. Around 'f(a)' the function will resemble the unique tangent that exists at that point which happens to be the derivative of the function at that point 'L(x)'. For the x value 'a' , f(a) = L(a), also with regard to the current situation, it appears as if f(a+∆x) = L(a+∆x), this only happens as long as ∆x is infinitesimally close to zero, and when dy = ∆y. Hence approximating the value of f(x) around f(a) using it's derivative L(x) and approximating the slope of the tangent like L(x) using values around f(a) is possible. Based on this we can conclude that the ratio of (dy/dx) does exist at a certain point f(a) when dy = ∆y.
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.
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.
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
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.
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)
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
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"
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.
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 ❤
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 ?
"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
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...
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.
are there any bad differentiable functions for which the differential dy is not a good approximation of their change Δy, no matter how small Δx we pick?
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.
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.
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?
I remember a Mathematics graduate talking about "infinitesimals" when explaining this to me. Fortunately I read an economics book that had this same explanation. Im glad to see it here again. You describe dy as an approximation of delta. But can we say delta y/delta x is an approximation of dy/dx for small values of delta x, right ?
Hello , I’m so curious to know what is (i)! , I reached to two integrals from 0 to infinity of e^-t *cos(lnt) dt and the second e^-t *sin(lnt) dt . but I couldn’t continue , could you do I video for that? Thanks 🙏
Factorial is only defined for positive integers, and extended using the gamma function. I think the backgroudn for the calculation of Gamma(i) is out of the scope of this chanel, but if you want someone to help you with that you should probably ask about it on math stack exchange or a similar site
Sebastian Gudiño I found the solution , my problem wasn’t in gamma function, my problem was just these integrals and then I’ll be done . to solve theses integrals unfortunately I couldn’t solve it , so I used wolpharm alpha. Thanks alot!🙏
Why did we assume that dx=Δx? Is it an exact value or a very good approximation? I am kinda unfamiliar with the use of differentials. I have only used them in integrals, and i guess in differentiation, although i have only just started using the dy/dx notation.
Is there a dy definition without using derivative. Approximations can also be achieved by regression. Is this the meaning of your remark regarding derivative?
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|
D means derivative, and in your example, usually people would put (d/(dx)) in front of y so it's like d/(dx) y, meaning derivative of y but in respect to x.
Todos los videos están muy interesantes. Hace una semana descubrí esta página y desde entonces he visto varios videos y todos me han gustado. Muchas gracias. Tengo algunas preguntas, sobre temas de matemáticas que no puedo hacer por este medio, si tienes un correo donde pueda enviar archivos y/o imágenes, agradecería me lo enviaras. Soy profesor retirado pero sigo estando interesado en aprender y en esta pagina he aprendido mucho. Felicitaciones blackpenredpen y gracias de nuevo.
Este comentario esta bastante fuera de lugar en un canal en ingles manejado por un asiatico con chino como lengua nativa. Ademas pedir informacion personal de esa forma no es una practica que se suela llevar en un medio como youtube que se basa en el concepto de comunidad. Si tienes interes en hacer preguntas sobre matematicas que sean respondidas por gente con conocimiento de la materia un sitio como Math Stack Exchange puede ser ideal para ti. Pero es un sitio donde se habla en ingles, y tu comentario en español, me hace creer que no dominas ese lenguaje. Pero estoy seguro de que debe existir algun sitio similar en español. Una busqueda de google te podria dar buenos resultados
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.
@@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.
Thank you for the great content. I have got one question please: given dy/dx=1/(2*sqrt(x)) then why do we have to multiply 1/(2*sqrt(x)) by dx if we would like to calculate the integral of dy/dx such : ∫ dy/dx = ∫ [1/(2*sqrt(x))] * dx ? I hope to get an answer :)
If you're doing rigorous real analysis then dy=f'(x) dx means "Delta y is asymptotic to f'(x)*Delta x as Delta x approaches 0", and if you're not doing rigorous real analysis then dx and dy might be infinitesimals. But under no circumstance are dx and dy plain old real numbers. "Let dx = -0.002" is like saying "Let 5 = 3". [edit] or maybe like saying "let rectangle (0,0)(0,3)(1,3)(1,0) = 3". It's not a variable, and even if it was it's not the same kind of object.
And I'm not denying that your method gives the right answers. I would be perfectly happy if you'd said f'(x) = dy/dx ~= Delta y/Delta x Therefore Delta y ~= f'(x) Delta x
let y=f(x)... let delta-x be small change in x so that it results delta-y change in y. If lim delta-y tends to 0, it is dy. And if the delta-x u took is so small that it is near to 0, like 0.0001, then it is dx.
For those not aware, what he described is essentially Linear Approximation, or Linearization
ua-cam.com/video/XQIbn27dOjE/v-deo.html 💐
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
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.
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 =)
@Conor Zayden Thanks so much for your reply. I got to the site thru google and I'm trying it out now.
Takes a while so I will get back to you later with my results.
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.
> ∆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
Holy moly this was so much better than my professors explanation! He literally was trying to explain the concept using marble and bread.
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!
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.
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.
@@Davith_D spot on! In simpler terms as ∆x approaches 0, the values of dy and ∆y will appear to coincide, and this takes place at an infinitesimal level, where ∆y is the change in the value of function 'f(x)' and dy is the change in the value of the Tangent line/ derivative function 'L(x)'.This can be understood using the concept of local linearity. If you zoom closely into a function 'f(x)' which is a differentiable curve, the function tends to look and behave like a line around a certain point on the function 'f(a)'. Around 'f(a)' the function will resemble the unique tangent that exists at that point which happens to be the derivative of the function at that point 'L(x)'. For the x value 'a' , f(a) = L(a), also with regard to the current situation, it appears as if f(a+∆x) = L(a+∆x), this only happens as long as ∆x is infinitesimally close to zero, and when dy = ∆y. Hence approximating the value of f(x) around f(a) using it's derivative L(x) and approximating the slope of the tangent like L(x) using values around f(a) is possible. Based on this we can conclude that the ratio of (dy/dx) does exist at a certain point f(a) when dy = ∆y.
Just did a calc midterm and this video would've bumped me up from a 70% to an 80%. For the sake of future math tests, you will be my new professor!
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.
♥
ua-cam.com/video/XQIbn27dOjE/v-deo.html 💐
It was what I needed today.
I watched your video to learn actually but I suddenly smile when you're looking and smiling at the camera. How cute of you💕
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 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
This makes my concept clear now! Thanks
Yay!!
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
Thanks mate by far the easiest explanation vid I have found on UA-cam. You must be a great tutor 👍
Thank you very much! I am revisiting this topic in Calc 3 and now I get it
ua-cam.com/video/XQIbn27dOjE/v-deo.html 💐
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.
Great explanation, searched a lot but this one is amazing!
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)
dy is also a change like ∆y but this change is estimated value. Calculus can pull out very detailed information.
My boi is slaying the professors with a 10$ mic.
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
That cleared out some of my back of the mind doubts
WOW! This is one of the best videos on UA-cam.
ua-cam.com/video/XQIbn27dOjE/v-deo.html 💐
Can you please share what course are you studying. Your knowledge of calculus is soo coool
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"
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.
This is the best explanation ever! Thank you!
His geometric representation is wrong 😊 go and check mit Herbert gross lecture approximation
its very useful for me ,ive just confused by the class course ,when i saw this video ,it makes my brain clear again
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 ❤
Nice content man, it really helped
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
Your videos really remind me of just how much I've forgotten. :)
I’m stuck because I don’t have a calculator with me 😆
So, a great oversimplification of this explanation is that dy/dx are the limit where DELTA x -> 0 of DELTA y / DELTA x ?
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...
You’re a good teacher. Thanks
Thank you for this explanation. It really make me understand better
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
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.
are there any bad differentiable functions for which the differential dy is not a good approximation of their change Δy, no matter how small Δx we pick?
try beyond delta x = 5
thank you, you make it so easy to understand what exactly does "d"y mean.
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.
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!!
How will be the curve if the value of delta y is negative?
What if x2 taken so big say 449397, then dy become way higher than exact change in y (∆y) then how to write ∆y ≈ dy
Thank you Sir....
A student from India...
They skip this in class and say to memorize the formuals...
This is so dope
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?
finally i've understood the difference betwen derivatives and differential thaaaaaaanks
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?
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
Thanks so much for these videos
I can't understand math the way is teach it in school
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ANOTHER ASIAN MATHEMATICAL GENIUS
You explained the difference between delta y and dy. But what is the difference between delta x and dx?
I remember a Mathematics graduate talking about "infinitesimals" when explaining this to me. Fortunately I read an economics book that had this same explanation. Im glad to see it here again. You describe dy as an approximation of delta. But can we say delta y/delta x is an approximation of dy/dx for small values of delta x, right ?
@Falchion [SSBU] my mistake. I meant dy/dx
@Falchion [SSBU] Of course, we are talking here about small dx
@Falchion [SSBU] of course, but you can make dx as small as you like without being 0.
This video saved my life two times
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Excellent explanation! Thank you!
Hello , I’m so curious to know what is (i)! , I reached to two integrals from 0 to infinity of e^-t *cos(lnt) dt and the second e^-t *sin(lnt) dt . but I couldn’t continue , could you do I video for that? Thanks 🙏
Factorial is only defined for positive integers, and extended using the gamma function. I think the backgroudn for the calculation of Gamma(i) is out of the scope of this chanel, but if you want someone to help you with that you should probably ask about it on math stack exchange or a similar site
Sebastian Gudiño I found the solution , my problem wasn’t in gamma function, my problem was just these integrals and then I’ll be done . to solve theses integrals unfortunately I couldn’t solve it , so I used wolpharm alpha. Thanks alot!🙏
Thank you sir for your clear explanation
Do one with Sigma and Integral too!!
Loved the explanation, really helped me
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Why did we assume that dx=Δx? Is it an exact value or a very good approximation?
I am kinda unfamiliar with the use of differentials. I have only used them in integrals, and i guess in differentiation, although i have only just started using the dy/dx notation.
Typically you choose whatever change in x you want, so this part is exact. The associated change in y will be approximate.
Is there a dy definition without using derivative.
Approximations can also be achieved by regression. Is this the meaning of your remark regarding derivative?
Thanks for making my concepts more clear . DrRahul Rohtak
Great explanation! Great video.
Bravo!! Crystal clear explanation!!
thank you very much. now I actually understand what differentiation tells us.
I think it would be better to further include sth in Taylor expansion. For example, the 2nd order derivative and the concept of reminder
Excellent explanations.... 👌👌
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|
what if I pick x=sqrt(pi) ?? I dont understand
If delta y is the exact change then what are we bothering with dy? What's the point?
This video save my life
If dy is the approximated change in y, then what does the d mean in y(d/dx) since this is equal to dy/dx
D means derivative, and in your example, usually people would put (d/(dx)) in front of y so it's like d/(dx) y, meaning derivative of y but in respect to x.
I really appreciate your hardwork. Thank you.
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Can you please tell me how to graph |y-1|=|ln(-|x|+1)| PLEASE
Todos los videos están muy interesantes. Hace una semana descubrí esta página y desde entonces he visto varios videos y todos me han gustado. Muchas gracias.
Tengo algunas preguntas, sobre temas de matemáticas que no puedo hacer por este medio, si tienes un correo donde pueda enviar archivos y/o imágenes, agradecería me lo enviaras.
Soy profesor retirado pero sigo estando interesado en aprender y en esta pagina he aprendido mucho. Felicitaciones blackpenredpen y gracias de nuevo.
Este comentario esta bastante fuera de lugar en un canal en ingles manejado por un asiatico con chino como lengua nativa. Ademas pedir informacion personal de esa forma no es una practica que se suela llevar en un medio como youtube que se basa en el concepto de comunidad. Si tienes interes en hacer preguntas sobre matematicas que sean respondidas por gente con conocimiento de la materia un sitio como Math Stack Exchange puede ser ideal para ti. Pero es un sitio donde se habla en ingles, y tu comentario en español, me hace creer que no dominas ese lenguaje. Pero estoy seguro de que debe existir algun sitio similar en español. Una busqueda de google te podria dar buenos resultados
Can you add 3 or 5 exercises for us to try this and see for ourselves? Thank you.
U explaination very well u were very cool throughout the video
It's very very clear now. Thank you soooooooo much
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Where does the infinitesimal come into the story?
this is linearization
Which is the first two terms of the Taylor Series.
can you also compare and contrast with del y?
Why did u put x=4 and not 5 in the last??? Help!
It's like the difference between summation and integraiton
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.
Very concise, very understandable. Thanks.
thanks, you are my math god
Thank you for the great content. I have got one question please: given dy/dx=1/(2*sqrt(x)) then why do we have to multiply 1/(2*sqrt(x)) by dx if we would like to calculate the integral of dy/dx such : ∫ dy/dx = ∫ [1/(2*sqrt(x))] * dx ? I hope to get an answer :)
It's amazing your explanation.
Thanks
Do you have a source for those definitions of dx and dy? They look completely wrong to me.
?
dy=f'(x) dx, and you take dx as delta x, that's pretty much it.
If you're doing rigorous real analysis then dy=f'(x) dx means "Delta y is asymptotic to f'(x)*Delta x as Delta x approaches 0", and if you're not doing rigorous real analysis then dx and dy might be infinitesimals. But under no circumstance are dx and dy plain old real numbers. "Let dx = -0.002" is like saying "Let 5 = 3".
[edit] or maybe like saying "let rectangle (0,0)(0,3)(1,3)(1,0) = 3". It's not a variable, and even if it was it's not the same kind of object.
And I'm not denying that your method gives the right answers. I would be perfectly happy if you'd said
f'(x) = dy/dx ~= Delta y/Delta x
Therefore Delta y ~= f'(x) Delta x
let y=f(x)...
let delta-x be small change in x so that it results delta-y change in y.
If lim delta-y tends to 0, it is dy. And if the delta-x u took is so small that it is near to 0, like 0.0001, then it is dx.
wait, so the entire premise of linear approximation is to approximate a curve using its derivative?