I also thought so. It is just unbelievable this quality is free. If this can be free, all the non-free lectures all over the world would be kind of scam.
Dear first time calculus learners, Do NOT expect to understand calculus after one pass through this video series. You must "pause and ponder" a lot, draw pictures, and see what new formulas you can discover through geometry. Read your textbook, listen to lecture, and do your homework problems, and make sure to give the sections in this video a pass or two or three (or twenty). Calculus is amazing and wonderfully intuitive, but was not invented in an afternoon, and there's a reason that the course is two semesters. If you fully understand these videos and can do computations and solve word problems, it is safe to say you have mastery over the material. Good luck and enjoy learning this beautiful subject!
+SybaPhoenix Gaming r/iamverysmart Someone could learn the basics of calculus from this video, but realistically, I doubt anyone could apply it without undergoing a more rigorous program. This is more of a supplement; something to make the subject click. Also, I've gotten an IQ of 150 on an online IQ test from choosing random answers. Online IQ tests don't count, as Danny Pike said. Proof: ua-cam.com/video/-r2n_mK9agY/v-deo.html
We live in an age where a highly motivated individual (with internet access and time) could learn just about anything with no formal education. I hope this playlist stays available for a long time because it clarified so many things I wondered about and couldn’t articulate.
Agreed. But I will say that formal education will continue to be a thing for quite some time since some people learn better with structure and an enforced routine. When I spent 2yrs teaching myself AI and coding, the hardest part of it all was keeping myself from going down all sorts of rabbit holes and various tangents to the point of not maintaining a central focus (as well as different sources perhaps speaking a slightly different "language" to describe things). Knowledge is so vast that no one can become an expert in it all, so reigns are needed in some form. What formal education allows for is teachers/professors having crafted a cohesive curriculum for each course, and academic counselors having aided a student in crafting a development plan to have that student's series of courses cohesively build towards a particular intellectual goal. I think what the future can bring, given this vast access to knowledge, is for people to get wiser as to how to map out a plan for themselves to most efficiently learn what it is they need to learn...since otherwise one runs the risk of learning a bunch of separate things but not particularly bringing it all together into a meaningful big picture. Videos like this are outstanding for piecemeal insight and learning, but one essentially has to be the master of their own ship in regards to how they'll want to apply that knowledge. It's the only guiding light towards them knowing what detail to dive into next. And like you said, that takes someone motivated - it takes quite a bit of discipline.
@@chrisjfox8715to me, the problem doesn't seem to be a lack of structure in the material of this lecture series but the lack of a clear method of internalizing and remembering the concepts (like schoolwork). Unless someone watching the video is taking a calculus course, there is no clear way for them to demonstrate and retain their new knowledge and reasoning skills (besides the occasional guided question). What this series does well is it teaches concepts in a way that is very clear, interesting, and motivating. The only thing it's missing (and what, in my opinion, many stem UA-cam channels are missing) is a way for you to remember what you learned. Give this series about 100 challenging, meaningful questions that connect to some curriculum, offer some incentive for people to complete them, and this series would be pretty close to perfect for someone to teach or reteach themselves introductory calculus.
@@kaboomgaming4255 Honestly, I (paradoxically) agree with _both_ statements, as someone who learned enough for a Calc I credit over about a year, but _still_ doesn't feel right now taking a Calc II class a year later.
Hi, I'm having a problem getting -x^-2. Can you please help me out? Area remains constant as we change the dimensions of the rectangle. Hence, initial area = final area. 1= [ x + dx ] [ (1/x) - d(1/x) ] 1= x (1/x) - x d(1/x) + (1/x) dx - dx d(1/x) x and 1/x multiply to give one. We subtract one on both sides to get zero on one side. We multiply both sides with -1. 0 =x d(1/x) - (1/x) dx + dx d(1/x) x d(1/x) + dx d(1/x) =(1/x) dx x + dx = (1/x) [ dx / d(1/x) ] x^2 + x dx = dx / d(1/x) As dx tends to 0, x dx also tends to 0, hence we can sort of ignore it. x^2 = dx / d(1/x) d(1/x) / dx = x^(-2)
**TIME-STAMPS TABLE** 0:06 Initial quotation 0:15 How to compute derivatives? 0:30 Why is such computation important? 0:45 It is abundant in real world 1:15 Important to always remember the fundamental definition of derivatives 1:45 D x^2 example 2:00 Graph analysis 2:40 Graphical intuition for Dx^2 3:30 dx^2 is negligibly tiny 4:20 Algebra passage to obtain derivative formula for x^2 4:45 D x^3 example 5:10 Delta volume of a cube 5:40 Negligibable parts 6:50 Pattern for Dx^a = a*x^(a-1) 7:30 Usually just symbols, but why? 8:10 We can ignore much of the terms in the computation 8:40 General case of x^2 and x^3 9:50 The importance of remembering the why 10:10 Example D 1/x 10:20 (You could just use the power rule) 10:45 Geometrical interpretation 12:00 Exercise for the viewer 12:30 Now figure out D sqrt(x) 12:40 Trigonometric functions 12:50 Geometrical view of trig functions 13:35 Starting by looking at the graph 14:10 D sinx should be cosine based on valleys and peaks, but why exactly? 15:30 Demonstration based on similar triangles 16:50 Now what is D cosx?
+Francisco Abusleme Also it takes about 30 minutes for me to make them, and they can potentially benefit more than 10.000 people (if 100.000 people watch the video and 1/10 needs them), so why not?
Oh my god. When I first saw this video at the start of college in my engineering course, I didn't have any clue how to solve the 1/x and the sqrt of x derivatives via geometric analogies. Now that I quit my engineering course and am pursuing a computer science degree, I finally solved it after 5 years. I finally figured out the tricks needed to solve both equations once I got comfortable with the concepts behind calculus. It was a roundabout journey for me. I know no one will read this, but I just wanted to share. It's a happy moment for me! Thank you 3b1b for this series.
A Small Doubt I derived them as told in the video. The Area Lost at Top = d(1/x)*x. -> d(1/x) Change in Height, x the previous Width The Area gained at Right = d(x)(1/x - d(1/x)) -> d(x) The Change in Width, (1/x - d(1/x)) -> The new Height And Intutively the Area Gained = Area Lost d(1/x)*x = d(x)(1/x - d(1/x)) d(1/x)/dx = (1/x - d(1/x))/x = 1/x^2 - d(1/x)/x We can leave out d(1/x)/x as this will a infinitisemally small = x^-2 ( But According to power rule it should be -x^-2) Can you please help on it
@@muthuraj3010 remember that if the area lost is the same as the area gained you can just work on one of them. Remember that you are always working to find the slope which is (y2-y1/x2-x1). let's call the difference in the areas h, and let's call y f(x). now you have that lim (h->0) of (f(x+h)-f(x))/h = f'(x). Plugging in the values you have ((1/x+h)-1/x)/h=f'(x), now apply the common denominator to get rid of the discontinuity in the denominator, (x-(x+h)/x(x+h))/h, now simplify and you are at -h/hx(x+h) --> -1/x(x+h) Finally apply the limit (h --> 0) so that you have -1/x^2 or -(1/x^2) which is equal to -(x^-2)=-x^-2
@@muthuraj3010 I don't know if you've worked it out already. But, in the first equation of "Area Lost" the d(1/x) should be negative as it is a decrease in the area. So, In my opinion the equation should be: -d(1/x)*x = d(x) (1/x - d(1/x))
this is actually how math should be teached, because for people who are interested in the topic this is just much easier to memorize. and i personally believe that people who are not interested in learning things like calculus should not be forced to, because it really helps noone if they are.
Math used to come really naturally to me, and after taking two years off from calculus on accident, I was struggling so hard and wondering how I even learned it the first time. I have been having those same moments like you mentioned and actually feel like I can do it!
So many educational videos on UA-cam are "edutainment" designed to give the illusion of learning something new, without actually teaching anything. This channel bucks that trend and I am SO grateful for it. Please never stop (or at least keep going for a really long time). Personally, I hope you eventually get into the math behind some concrete practical applications like machine learning algorithms, but I'm loving these pure math series too.
If you want to code one, a good starting point is this link: iamtrask.github.io/2015/07/12/basic-python-network/ I found it very useful in applying the theory of 3Blue1Brown to a real neural network. Happy coding!
the derivative graph of sinθ is literally mind blowing. two years of calc and it finally makes sense. thank u for giving me hope for my ap exam in a couple days, this content is incredible.
I remember taking the AP test. I got an A on the final test our teacher gave, but then I got the AP test and failed miserably. The AP test was much harder than anything my teacher threw at me.
@@singlemuskeeter6916 I am going to give something close to what you may call ap in you country. From most of the teacher's suggestion, its suggested trying an alternate solution for every question. And then pinpointing why did or why didn't that perspective work
I am a head of mathematics at a school in the UK and try my absolute best to teach my students and embed this sort of level of understanding. The one tool I just wish I had is animation! These animations are so clear. I use Geogebra to the best of my abilities but just can't quite offer the same visualisation as you do with these. What do you animate using? If I could just do animations a tenth as good I'd be happy. This level of visualisation adds that extra dimension for students to grasp a concept. I am very appreciative of your videos - once I have reached the limit to which I can explain something I show these videos in class to add that extra visual aid. So pleased to have your videos to complement my lessons.
I know this is a very late response, and I hope you're still the head of mathematics and my answer is still of use, but he uses a programming tool called manim in python.
*To whom have you directed your commands? You don't need to include two prepositions, regardless of whether you're speaking with overly-formalized English.
Next up will be "Visualizing the chain rule and product rule": ua-cam.com/video/YG15m2VwSjA/v-deo.html You’ll notice throughout this series that I encourage a more literal interpretation of terms like “dx” and “df” (aka differentials) than many other sources. I call this out and explain further in many of the videos, especially chapter 7 on limits, but given that students are often told not to take these terms too seriously, to be wary of treating them as literal variables, it’s probably worth adding another comment on the matter. The path between treating these terms as literal nudges and a fully rigorous treatment of calculus is actually quite short, considering the loose language that seems to be involved. You just need to understand two things that are implicit in the notation “df” and “dx”. First, the size of the nudge df is dependent on the size of dx. It is not its own free variable, and what it means depends on your current context. Second, for any equation written in terms of df and dx, when you replace dx with an actual number (e.g. 0.01), and replace df by whatever nudge to the output is caused by that choice of dx, the equation will probably be slightly wrong, with some error between the left-hand side and right-hand side. But what it means to be using these differential terms is that that error will approach 0 as your choice for dx approach 0. This is why terms which are initially proportional to (dx)^2, and hence retain a differential term even after dividing by dx, can be safely ignored. Even in the most rigorous proofs of derivative rules and properties, these tiny nudges show up, though often under the names "delta-x" or "h". The ideas presented here are essentially the hearts of those proofs but phrased without the surrounding formal language. I put together this series not just with calculus students in mind, but also with the hopes of pointing back to chapters here when I cover real analysis, the formal backbone of calculus, so I am motivated as much by an ultimate desire for people to understand the rigor as anyone else. (Also, as a hint to those asking about how you know that the triangles at the end are similar, use the fact that the tangent line of a circle is perpendicular to its radius.)
I made time-stamps table for this video and the last one, have you ever thought about adding them to your videos? They increase the value very much by making them also consultable rather than only watchable (people can come back to find a particular part)
Will this series primarily be based on Calculus 1 material or will the later videos cover Calc 2 and 3 as well? Fingers crossed for some awesome multivariable calc videos.
For the f(x) = √x case, the reason why the new area is represented by dx and not df (as in the x^2 and x^3 examples) is because we square both terms in f(x) = √x to get (f(x))^2 = x. The blue area is therefore f(x) * f(x), which is simply √x * √x = x. The new area, dx, is created by a 'nudge' df(x) in both directions, which is just d√x. From there dx = 2 * √x * d√x + (d√x)^2. Ignoring the (d√x)^2 terms since they go to 0, you get d√x/dx = 1/(2√x).
Hello! I am from Brazil and would like to thank you for your work, I am a student of Industrial Chemistry and in my country we have a bad basic education, at the time there were no platforms like this, preventing access to content like yours. Thank you so much for dedicating your time to the cause of education. It is very important to many people like me.
If anyone's wondering what the justification is for the claim he makes at 15:39: The base of the small triangle is perpendicular to the right side of the large triangle. The hypotenuse of the small triangle has a slope very close to the tangent of the circle at angle theta, and therefore is roughly perpendicular to the radius shown (the hypotenuse of the large triangle). Thus, the two angles of those two triangles that are touching are about the same. We also know they are both right triangles, so that's two angles that match. There's only one possible value left for the remaining angles (sum of interior angles of triangle = 180 degrees), so all the angles match, and therefore the two triangles are similar (well, mostly, but they get more similar for smaller values of d-theta).
I just started learning calculus. My math teacher taught me some formulas but when I asked him "but why?" he didn't really have an answer. Until I came across this channel I had many questions. I'm really loooking forward to next chapters. Keep it up.
The one time when I asked a teacher how the derivative formulas were, well.. derived, she told me to get a masters in maths. It was many years ago, and I kinda wish I had resources like these readily available back then. Maybe I wouldn't have spent over a decade avoiding everything to do with numbers, just because I was so jaded and confused.
@Tracchofyre that's why it is important that a teacher has sufficient mastery over the subject. To teach mathematics in high school you need to have a masters in mathematics, even though you won't use 90% of what you learned in university in teaching in a high school. But if you get too strict the supply won't meet the demand. You need a certain percentage of math, physics, etc teachers and the most talented students won't want to become a high school teacher, but they are exactly the people who can provide awnsers beyond an 'it is so because it is so' level.
Watching this series has really made me wish 16yr old me was as motivated and appreciative then as I am now at 33 of how interconnected the various maths are. I literally had a flash back to highschool and had a legitimate "ahaa" moment. This is truly excellent content!
I'm 17 and I have a great maths teacher but we haven't done derivatives yet but I'm writing a paper on the Fourier transform and I get lost very quickly and this has helped me so much with what I'm writing. Seriously, he makes great content, even with little to no knowledge about semi-advandced calculus, i understand all of it and its so great!!!
Im sixteen and just want to say that i very grateful to be sixteen and can comment on your comment ( although im not a native speaker so that simple comment have 300 erros. Ps : salve do brazil a educaçao aqui é uma merda. )
I took calculus almost 6 years ago now. I'm now a grad student in robotics and diffeq is life. I love seeing how some of these things come about that either: were never explained to me or had been forgotten due to the years of plugging away.
Hemanth Kumar I think most engineering fields use calculus, and robotics would be a category of mechanical engineering. So robotics SHOULD require it too. (Don't quote me on that. I'm not in robotics.)
It's not really just robotics. It's more like anything that is related to movement (change of position) requires the use of differential equations. Differential equations are used in most engineering fields and economics. Essentially anything to do with a rate of change can be represented by a differential equation.
I'm a math major currently finishing up my second semester of Advanced (i.e. proof-based) Calculus. I just learned more about why D sin(x) is cos(x) than in all my years of math up to now.
Reuben French as a math major, what do you think about disregarding dx raised to a power? Imo it is not rigorous and well defined to just disregard a dx if we're considering derivatives in this way..
@@sergioh5515You factor dx out of everything and can then divide by dx. You then evaluate it at the limit as dx approaches 0, so anything with a dx left (i.e. initially had dx to a higher power than 1) is multiplied by zero.
For people wondering how d (cos θ) = - sin θ Note: While moving around the circle, sin θ is increasing but cos θ decreases from 1 to 0 and then continues its simple harmonic motion. Just use that line of reasoning and you can see at 16:56 that derivative of cos θ is - sin θ.
I understand that the cosine decreases as ∅ increases. But this is only true for the first 2 quadrants. What stops us from making the same analysis on the last 2 quadrants and finding a relationship where cosine increases with ∅. The geometry of the problem would be the same. This would mean d(cos(∅))/d∅ = sin(∅) wouldn't it?
Here is my solution to 12:21 Area gained + Area lost = 0 Area gained = (1/x - d(1/x))*dx Area lost = x*d(1/x) Adding the areas x*d(1/x) + (1/x - d(1/x))*dx = 0 "Distribute" the dx x*d(1/x) + (1/x)*dx - d(1/x)*dx = 0 Rearrange to factor out d(1/x) in next step x*d(1/x) - d(1/x)*dx + (1/x)*dx = 0 Factor out d(1/x) d(1/x)*(x - dx) + (1/x)dx = 0 Subtract (1/x)dx from both sides d(1/x)*(x - dx) = -(1/x)*dx Divide both sides by (x - dx) AND dx d(1/x)/dx = -(1/x)/(x - dx) Distribute the terms in the denominator on the right hand side d(1/x)/dx = -(1/(x^2 - x*dx) The second term in the denominator on the right hand side will go to zero as dx goes to zero. The solution is: d(1/x)/dx = -(1/(x^2))
Can you explain why x*dx goes to zero (the last step)? I understand the whole point is that dx goes to zero but couldnt we just do it right at the beginning? Thanks a lot!
I think he wanted people to reason about it geometrically. Same goes for the root. Am I wrong? Then again there's no way to write a geometric solution in the comments)
For the case f(x) = 1/x: The blue area + red area (area lost) = 1 The blue area + green area (area gained) = 1 This implies the red area (area lost) = green area (area gained) Red area = -d(1/x) * x Green area = [(1/x) - (-d(1/x))] * dx = [(1/x) + d(1/x)] * dx Since red area = green area, we have: -d(1/x) * x = [(1/x) + d(1/x)] * dx Dividing both sides by x * dx, we get: -d(1/x)/dx = [(1/x) + d(1/x)] / x Ignoring d(1/x) on the right side since it approaches 0, we have: -d(1/x)/dx = (1/x) / x -d(1/x)/dx = 1/(x^2) Dividing both sides by -1, we get: d(1/x)/dx = -1/(x^2) Therefore, the derivative of 1/x is -1/(x^2). Power rule d/dx(x^n) = n*x^(n-1) works even when n = -1.
very nice, but I don't understand why d(1/x) should be the one approaching 0, weren't we seeing what would happen as dx->0 and looking at d(1/x)? Because otherwise d(1/x) won't get cancelled from the right side.
@@eriksolis6176 I have a question. Why is it 1 = [ x + d(x) ] [ 1/x + d(1/x) ] and not 1 = [ x + d(x) ] [ 1/x - d(1/x) ] (due to the fact that the 1/x-d(1/x) side of the rectangle is getting smaller )? thank you
@@shiluka I don 't quit understand. if we calculate the limit of x when it's aproaching 0 the lim =0 however if we calculate the lim1/x when it aprroaches 0 it will be +∞ (plus infini) so (what i think ) if As dx → 0, the term d(1/x) on the right side becomes very big compared to the other terms. Am i correct ?
Nathan Richan In the limit as dtheata goes to 0, the side of the small triangle on the circle will be perpendicular to the hypotenuse of the larger triangle. You can use this fact with corresponding and alternate angles to see that the internal angles of the two triangles must match. Thus they are similar
Gregory House '' Then, because the angle between both opposites sides of both triangles with respect to θ is 90°, then the other angle on the new triangle must also be φ. '' Dafuq ?? how is the opposite side to θ of the triangle making 90° with the opposite of θ of the new is supposed to mean anything ? I mean you can have a completely different triangle having this exact same property
For those who are not understanding this, just keep rewatching this video and do not give up. Even Im going for a 4th rewatch and now it seems that im starting to appreciate its beauty!!
same, I'm watching this video for the 4th time as well, and I've watched this video 4 years ago. I CAN FINALLY FIGURE OUT WHAT HE'S TEACHING. It actually takes a little dive into calculus beforehand in order to fully understand this video and the entire series.
@@Willy_Wankaonly people who think they are stupid and want to feel like they are smart undermine others, you do it to feel better about yourself, that doesn't make you any smarter.
I graduated in the 90's with a BS in math and did not get beyond slope is derivative and area is integration. Man this video and others by this channel bring on a strong feeling of regret from missing how much knowledge was missing below the waterline of the calculus iceberg. Do you feel me?
I am a Vietnamese student, I can remember lots of derivatives but never did I understand their meanings. But only until I find out this channel, it's enlightening!
15:50 the reason that "little angle" is equal to θ is because the hypotenuse of the small triangle is considered a straight line, and therefore it can be considered the TANGENT of the circle. Since it is the tangent, it is perpendicular to the radius of the circle, and the rest is now obvious.
I understand why we would consider the arc line as the hypothenuse of a triangle, but still don't understand why the triangles are similar. Why is theta back here and not another random angle ?
12:27 solution Note - sqrt(x) means (root x) i.e. (x)^(1/2) to find - (d sqrt(x)/dx) dx=new area dx = sum of the areas of two rectangular strips + area of small block dx= 2 sqrt(x).d sqrt(x) + d^2 sqrt(x) here d^2 sqrt(x) can be neglected as has power more than one dx= 2 sqrt(x).d sqrt(x) 1/2 sqrt(x) = d sqrt(x)/dx hence solved.
@@cheva1 Since you are squaring a product, you essentially "distribute the exponent" so that it becomes d^2(x). Good catch. I didn't notice that at first.
He’s explained so many math concepts better than any teacher or professor that I had. I took calc 1 and 2 but never was able to fully grasp what derivatives are, how they work. This video did explain it so well.
I'm an ex-gifted kid, probably neurodiveregent, I used to like maths at school, but now as a university student feel like I'm failing everything. Watching this series helps me both objectively understand calculus better and subjectively feel less paralyzed by self-hatred. The animations are so smooth and they keep my attention from hopping over to something else. Every video has subtitles. The whole series emphasizes understanding over memorization. What I'm trying to point out is that it's not only generally awesome, it's also pretty much neurodivergent-friendly. Don't know if it was intentional or not, but thank you in any case.
This is all so simple yet so profound. I love rediscovering calculus through non-hostile eyes. the whole animation involving 1/x was so elegant i loved it
This is a monumental achievement. When I was young I was exactly like the author who wants to understand every equation by visualizing how it works, only to be put off by incompetent teachers and pathetic syllabus which only cram you with queer equations without ever bothering to explain them. Now with so many maths explanation videos out there, I have to say 3Blue1Brown is the most concise and elegant one I have seen. I don't think saying the author has done a huge contribution to human civiliaztion would be an overstatement.
Before I bumped into your channel, I had almost only algebraic intuition than the visual side. In order to make the algebraic process get etched into my intuition, I imagine that I was to explain those math concepts to some family members who were conventionally deemed as ‘have no mathy brains’, such as my brother, whose highest diploma is from primary school. And the reasoning process needs to be as plain as possible so that it fits Einstein's instruction to us: ‘If you can't explain it simply, you don't understand it well enough.’ This visual math induction of yours is somewhat like a superpower to me. And thinking about it like a superpower makes me wanna learn it. So I made watching your videos part of my morning routine. The surprising result is that now I can confidently say these two things: Math is fun. Getting to know math is NOT that intimidating. Thank you. Grant.
I literally have never seen(heard actually) a better teacher than you. You are actually helping us students alot by making these videos. I hope something really good happens to you someday.
I've been applying the Power Rule so many times ever since I learned about derivatives in calc, but never truly understood why the formula is the way it is. After seeing the geometric visualizations for x^2 and x^3, it makes a lot of sense now. Thank you for making these videos, seeing all these different interpretations of formulas I didn't give a second thought about is really enlightening. I look forward to the next 7 days of videos.
Oh geez I did the viewer challenge! For once I actually completed a viewer challenge! I know people are gonna think I'm dumb for finding that "breakthrough" profound, but I did a viewer challenge!
Best tutor everrrrrrrrr I am a Biology Olympiad participant and I needed a good comprehension of derivatives and integral for statistics, population ecology, probability and physiology topics which I accomplished with this channel's videos. Thanks a lot. edit: I'm Iranian and I'm aware of the lack of fluency of English and accessibility to UA-cam among Iranian students. I would be grateful if you give me the right and cooperate with me, so I can translate your tutorials and share them with my friends.
Holy balls this series. I had one math teacher in school that taught us new formulas by going over how they are actually developed and I could understand everything perfectly, but after he left to teach at a university, I never understood what I was doing. I was just remembering how to use formulas. You have no idea how useful all of this is to me... I've already taken calculus and got ~65% in it, and it's part of my University course so I get to take it again and I've watched 2 of these and I already feel like it's going to be a piece of cake. If you are still around when I get a real job out of this course, I will repay my debt to you :))
@@SuperYtc1 Lol, got 95% in that calculus course and the one the following year, can't get a job 👍 gotta love going back to these cringy comments from years and years ago
@@DreadJester448 Congratulations on the 95%! I also did maths at a good uni 4 years ago and still have no job. I am learning about programming now and want to get into AI, hence refreshing calculus and this course is great. Getting a job is tough especially with all the stupid employers around. :( Goodluck!
I'm about 50 yo, all my life I was afraid of math. Calculus was a nightmare for me. With this channel, I feel like I've defeated my ancient fears. Thank you
For those who are struggling with 12:15 d/dx (1/x) If you try to solve this the following way, you will get the WRONG ANSWER xy = 1 (x + dx)( y - dy) = xy = 1 This is because, you have ignored the fact that dy is already negative and there is no need to put another negative sign in the (y - dy) term So the real method becomes (x+dx)(y+dy) = 1 xy + x(dy) + y(dx) + (dx)(dy) = 1 Since xy = 1, and dx and dy are both tiny so their product will be negligible x(dy) + y(dx) = 0 x(dy) = -y(dx) dy/dx = -y/x Since y = 1/x d(1/x)/dx = -(1/x)/x d/dx (1/x) = -1/x²
According to the video i write it this way : df = 1 - 1 = 0 = (x +dx)*1/x - x*(1/x-d(1/x)) U juste calculate and get the answer Stay geometric and u look d(1/x) as being a distance (so positive) and use - signe when needed For the square root it's even simplier : just use the same drawing as for x square (replacing both x on x and y axes by √x on both) and u get the answer as (x)' = 1
correct me if I am wrong, "x*d(1/x)+1/x*dx=0", right? "x*d(1/x)" is the area lost and "1/x*dx" is the area gained but the area lost and the area gained are the same, so we put "0" on the right side.
How do you get from this expression: x * (dy) = -y * (dx) this expression: dy / dx = -y / x ? It is equivalent as if from this expression: x * a = -y * b you get this expression: a / b = -y / x right? How did you derived it?
15:39 why the triangles are similar (commenting so i can look back at this, except im not a big brain math genius like everyone else here) - big triangle angles: θ + 90°+ (other angle)= 180°, so θ + (other angle) = 90° - radius/big t's hypotenuse is perpendicular to tangent line of circle (hypotenuse of small triangle) - knowing that alternate interior angles are congruent, angle btwn radius and bottom part of small t is θ - because angle btwn tangent line and radius is 90° (hypotenuse of small and big triangle), 90°- θ = (other angle) - this means that the far right angle of small t is "(other angle)" - because small t has a right angle and has (other angle), and θ + (other angle) = 90°, the last angle is θ. - because the angles of both triangles are the same, they're similar
you are math god. It took me years of study and even more research to understand the essence of math. Wish u existed 10 years ago :(. I had only one good math teacher in collage, but u outshine everyone. Your explanations are simply beautyful, intuitive and simple. When i was studying i had the same approach to math problems. PLEASE PLEASE continue your work. I would like to see you explain FUNDAMENTAL FORMS, FOURIER SERIES AND SPHERICAL HARMONICS. I had very hard time to understand those. I consulted countless professors and used Bronstein math manual, wolfram wiki, everything. Still those are still abstract subjects to me. Pls help
I am thirteen and now I'm here listening to 3B1B talk about the essence of calculus. He truly simplified the complex and 'boring' ideas into some simple but beautiful, wonderful, and interesting drawings. Best respect.
What I think is that instead of turning boring things into interesting things, he shows how beautiful maths is, as what it actually is. Maths is always attracting me, like how bread attracts a starving person.
3:58 Maybe a better explanation than "this is so tiny, you can ignore it (nevermind the other term also gets truly tiny and will not be ignored)" would be to actually divide by dx once to get df/dx=2x+dx and let dx approach zero, so that df/dx approaches 2x. EDIT: You did actually explain it this way at 6:11 :)
This really helped clarify, but could you help explain why area gained shouldn't equal area lost? Wouldn't this necessarily mean then that Area gained - area lost = 0, meaning that the total area remains at 1?
Subtle tweak to make (6 months late). Area gained is actually (1/x - d(1/x))*dx. Expansion gives you a dx*d(1/x) term. When you divide by dx, you are still left with a d(1/x) term which will still be really small, so we just forget about it (as with dx terms in previous examples). The answer ends up being the same, but this way clarifies the similarity between this and the x^2 case :)
Đạt Trần Because we're defining the box to have an area of 1. When we add on another section, we know that the area removed must be the same so that the area of the box stays the same.
for those wondering about the derivative of 1/x here's what I did : the text stated that the area lost must be equal to the area gained in terms of value so theu must be of opposite signs : x *d(1/x) = - (1/x)dx >> d(1/x) = -(1/x^2)dx >> df/dx = - 1/x^2 (where f = 1/x) hope this is a logical reasoning
I’d love to say a huge thank you to you. All the videos you have made are absolutely fascinating and beautiful. I remember being so deeply moved by maths when I first saw your topology videos. They have motivated me a lot to pursue mathematics in my further studies and I am so glad to have you to be my best maths teacher. Don’t stop making videos and thank you very very much!!!
Here's what I got for the f(x) = 1/x problem. Looking at the small rectangle with sides dx and d(1/x), we know that the derivative is the ratio of its height over width (its slope) as dx approaches zero. Using the graph, we find the width = x + dx - x = dx, and (remember to substitute in x + dx) the height = 1/(x + dx) - 1/x = (x - x - dx)/(x*[x + dx]) = -dx/(x^2 + x*dx). So then the slope = (-dx/[x^2 + x*dx])/dx = -1/(x^2 + x*dx). Now as dx shrinks to 0, so does the x*dx term in the denominator, and we are left with -1/(x^2).
Here's an alternative solution: We know that the area lost equals the area gained, so we can make the equation: -x * d(1/x) = dx * 1/x ^ (important note: the left side, being removed, is negative. this is just a rearrangement of x * d(1/x) + dx * 1/x = 0) Now, remember that our goal is to find an equation that will leave us with: d(1/x)/dx = some value. Let's try rearranging the above equation to get to that point: -x * d(1/x) = dx * 1/x 1. divide both sides by -x - note that this is *not* addition so you don't need to distribute the division. You may think of it as multiplying both sides by 1/-x = d(1/x) = dx * (1/x) * 1/-x 2. divide both sides by dx = d(1/x)/dx = (1/x) * (1/-x) 3. simplify = d(1/x)/dx = -1/x^2 4. we've found the answer!
@@NateLevin "-x * d(1/x) = dx * 1/x" Why is only x negative here? How is it negative? "note that this is not addition so you don't need to distribute the division." What does this mean?
@@qleo1769 If you look at the red and green part, the area lost is equal to the area gain, the loss of area represents that negative notation at the left side of this equation
I tried to get it using the nice equation in the first episodes. A small change in y divided by a small change in x. I got (sqrt(x+dx)-sqrtx)dx. I tried thinking of ways to simplify it, and nothing looked like ti would work until I set it equal to another variable, which I called y. I got x + dx = dx^2 y^2 +2dxysqrtx +x. Miraculously, the x cancelled out and left me with dx=(dx)^2*y^2+2dxysqrtx, which is incredibly easy to solve, and so left me with -1/(x^2)
In the czech language, the word education comes from farming, where farmers would "educate" the land - prepare it for actually growing (in work, but also life in general). I think this series is exactly what education is supposed to be. Thank you!
Grant, Australian viewer here As an aspiring physicist and mathematician, I’ve always loved these vids. Recently I was invited to participate in our schools year 10 accelerated maths course (currently year 9) and so to prepare I’ve been self-teaching some introductory calculus. These videos are just sublime, I’m learning so much and can now almost consistently replicate your results, always using the visual interpretations. The videos are legitimately going to change my life and career, so thank you so much for this amazing content!!
This video was sooooo satisfying to watch, especially considering how when I was learning calculus, I'd ask, "Wait, why does this differentiation rule work?" and my teacher would say, "Oh, it's just derived from the limit definition of a derivative."
mollymack98 Well, it is. None of those rules are defined by pretty pictures, but visualization helps us to build a mental heueristic, which enables us to find proofs.
For the challenge at 12:27, I propose the following solution: d(x) is the new area (i.e the yellow area) That mean we got: d(x) = d(√x) √x + d(√x) √x + (d(√x))² Which we can bring to: d(x) = d(√x) (√x + √x) + (d(√x))² If we divide both sides by d(√x): d(x)/d(√x) = 2√x + d(√x) If we take the inverse of both sides we get: d(√x)/d(x) = 1/(2√x + d(√x)) And as d(√x) tends to zero it becomes negligible and we finally get: d(√x)/d(x) = 1/2√x Which is the derivative of √x, hope that helps.
Thank you for your comment! You helped me connect the dots backwards. I guess I was biased by how the input “x” was “length” in previous examples, but here it’s area. The challenge I faced was in verbalizing the function to aid its visualization… x^2: gives area of square of length x; 1/x: gives height of rectangle of length x such that area is 1; x^(1/2): gives length of square of AREA x;
Help me understand, why did we take the inverse? I understand it's to satisfy the "d√x/dx" part, but why does that have to be the form of the answer in the first place?
@@cendolgbf remember, each time u calculate the derivative u gotta divide the tiny vertical nudge which is d√x in this case by the tiny horizontal nudge which is dx in every case, moreover if u look throughout the video he always calculates the derivative of a function f with df/dx, so remember the tiny horizontal nudge which is dx is always in the denominator, that's why I took the inverse, it's not just cuz it's a useful trick I pulled out of nowhere, hope that helps.
@@azizautop995 right, i was confused by the subtle change that d(x) is now the area instead of the length (in other examples). That makes a lot of sense now, thanks!
I cannot stress enough how helpfull this has been. Going through Highschool and Uni where only surface level explanations are given can dissolution you and make you forget why you ever liked math and science in the first place. These videos are helping so much to re-ignite my curiosity and remind me why math and science exited me so much in the first place.
12:21 If you zoom in the point the ratio d(1/x) / dx (height of that small rectangle/ width of the small rectangle ) should be the same as the ratio of the bigger rectangle (1/x) / x hence it is 1/x^2 and ofcourse the sign is because d(1/x) is actually negative. -d(1/x) / dx = (1/x)/x d(1/x) / dx = -1/x^2
Maybe you guys can think like this way easier: In order to keep the area equals to 1 all the time, the new amount of gain【(1/x)*dx 】must equals the area of loss【x*d(1/x)】 when the tiny x change. This gives us the equation(1/x)*dx=(x*d(1/x), then we can get new equation d(1/x) / dx=(-1)*1/x^2【since the nominator change is negative we need to times -1】
@@ming5363 DISCLAIMER: assume dx in infinitely small, so an incredibly small increase of dx would yield an incredibly small decrease away from our 1/x, which would practically retain that original side length. So 1/dx would approximately be = to 1/x
Here is my best analysis d(1/x) = df He says df is a negative length, ok x*df + (1/x + df)*dx = 0, after dx occurs the height of rectangle to the right is not exactly 1/x anymore it reduces by df. remember, df is a negative length in diagram above. -x*df = dx/x + df*dx df*dx is negligible and essentially zero, cancel -x*df = dx/x df/dx = -1/x^2
@Sotobito Well no.. The area of gain is 【(1/x)*dx 】. Why are you trying to add the area of that little transparent box on top of green filled rectangle? That green filled rectangle is what represents the area of gain and its equation is 【(1/x)*dx 】only
The Solution at 12:16 Since the area should remain constant that is 1unit. Therefore area of rectangle (x+dx)*(1/x-d(1/x))=1, x*1/x-x*d(1/x)+dx/x-dx*d(1/x)=1, since dx*d(1/x) is very very tiny value we neglect it , therefore x*1/x-x*d(1/x)+dx/x=1, 1-x*d(1/x)+dx/x=1, x*d(1/x)=dx/x, d(1/x)/dx=1/x^2. Since, d(1/x) decreasing the height of the rectangle we take symbol to be negative.
This is so good. My second time watching and this time taking notes and drawing some of the diagrams. I am so grateful for you sharing your experience.
Your videos are so beautiful. They really express the pure beauty and elegance of mathematics and also physical phenomena. Unfortunately, not everyone on the earth can or wants to experience this beauty. I am privileged. Thanks !
Hi, I'm having a problem getting -x^-2. Can you please help me out? Area remains constant as we change the dimensions of the rectangle. Hence, initial area = final area. 1= [ x + dx ] [ (1/x) - d(1/x) ] 1= x (1/x) - x d(1/x) + (1/x) dx - dx d(1/x) x and 1/x multiply to give one. We subtract one on both sides to get zero on one side. We multiply both sides with -1. 0 =x d(1/x) - (1/x) dx + dx d(1/x) x d(1/x) + dx d(1/x) =(1/x) dx x + dx = (1/x) [ dx / d(1/x) ] x^2 + x dx = dx / d(1/x) As dx tends to 0, x dx also tends to 0, hence we can sort of ignore it. x^2 = dx / d(1/x) d(1/x) / dx = x^(-2)
I'm 14 years old and learned how to do calculus only by watching this videos and practising with example tasks/playing around with the thoughts. Great series, great channel, great explained animated math! I suggested it to all my friends. Thank you 3blue1brown!
It’s imperative that 3B1B is kept on the internet forever, for everyone. The value of this channel cannot be quantified. I’ve learned so much from Grant.
Key to find the solution is: searching for a conncection between dx and d(f(x)) and if necessary cancel out the parts, which get smaller and smaller for a tiny dx: Solution for 11:31: The derivative of the function f(x) = 1/x is to be sarched: The visulation shows the input as the length of the rectangle and the output as its hight, while the area stays 1. You recognize that the area removed has to be the same as the area added: Area removed: x*-d(1/x) Area added dx * (1/x) x*-d(1/x) = dx * (1/x) /// divide dx, divide x - d(1/x) / dx = (1/x) * (1/x) d(1/x) / dx = -(1/x) * (1/x) = -x to the power of -2 = the derivative of 1/x Solution for 12:28: The derivative of the function f(x) = sqrt(x) is to be searched: This visulation shows the input as the area of the square and the output as its length. dx is the added area; d(sqrt(x)) is the added length; The added area dx can be expressed with d(sqrt(x)): dx = 2 * d(sqrt(x))*sqrt(x) + d(d(sqrt(x)) squared The squared edge arrives 0 for smaller and smaller dx -> can be canceld out; All we need to do is, solve the equation: dx = 2 * d(sqrt(x))*sqrt(x) /// divide dx 1 = (2 * d(sqrt(x)) *sqrt(x)) / dx /// faktor 1/2; factor 1/sqrt(x) 0,5 * (1/sqrt(x)) = d(sqrt(x)) / dx = the derivative of sqrt(x)
Thanks, but I don't understand how we choose which variable represents the area for example in the visualisations, sure they work for these examples but I have no idea how to do them myself when sometimes dx represents the length added and other times the area..
I'm from China, and my words just fails to express my gratitude to 3B1B! I'm a student from data science, and I really needed it to refresh my memory ! Thanks a lot!
I'm only in seventh grade, and though it takes me a lot of time to understand this stuff and probably need way more learning in the areas of linear algebra, this guy does a great job helping me understand core conceptual ideas of a part of math I won't be taught until high school
Hey, Professor Bertrand! For the first homework exercise, I found the answer by adding the areas of the rectangles (the red and the green), setting that sum equal to 0 and solving for d(1/x)/dx. For the second homework exercise, I found the answer by using the same idea as with the square with side length x, replacing x with sqrt(x), and then solving for d(sqrt(x))/dx. Finally, I compared these answers to the answers that would have been expected from using the power rule, and found that they were exactly the same. Thank you so much! I really enjoyed this video! Keep up the good work!
Your videos have taught me to imagine a lot. I'm an aspiring data scientist and many of my friends follow your content. A request will be making such short series of probability and statistics series. Would be really helpful.
Watching these videos I can see how Math education in schools suck really really bad! Why the hell on earth we didn't have such methods and teachers to teach us Math. I could still use a teacher who knows math by heart and can teach even this is my junior year in University. I hope people who memorises Math and think they know Math, stop being teachers and leave some space for teachers like this.
dunno, but i would probably struggle to understand things like this when they are presented only geometrically. So i doubt i would understand what it is all about. for example, why triangles presented are similiar?
The difficulty with always teaching conceptually/intuitively is often time, a packed curriculum, and students who come in with a poor understanding of pre-requisite material. I teach calc I, and focus on intuitive aspects as much as possible, but time is always against me, especially when students struggle with pre-requisite material and, as a result, struggle to see some of the beautiful underlying calculus.
Cowmoo83 Yes, you are right with the time issues. The teachers who teach pre-requisites should be better so that you could be better as well. But I think it's not a simple problem to solve. There are some pretty good approaches with education. For example Sal Khan's learn or repeat approach as I call it, is a good one. You can find it on his TED talk. I myself, believe that as humanity we are well below our potential. To reveal that potential, we need superior education for everyone.
12:21 We have that xdf + dx(1/x - df)=0 since the area remains constant. I isolate df/dx with simple algebra obtaining df/dx= -1/(x^2) - df and since df is negligibly small we can cancel it.
For the other we have that dt^2/dt is equal to 2t so i just take the reciprocal of both parts and substitute t=sqrtx so i have that the derivative is (2sqrtx)^-1.
Somehow the very first step I did was already wrong. My initial equation is: x * 1/x = x * 1/x + dx * (1/x - d(1/x)) - x * d(1/x) Basically saying: The area = the area + the stuff on the right - the stuff on the top. I guess I have to keep d(1/x) negative even in the "stuff on the top" part. For anybody wondering, I gave up once I had: d(1/x) / dx = 1/(d(x) * x) + 1/(x²)
If the area remains constant, doesn't that mean that xdf = dx(1/x-df) ? I get df/dx=1/x^2 - df/x , so the opposite sign. Where am I wrong with my logic?
Well done! I have finished my first half semester in Calculus 1 at a private University and haven't learned as much in 2 months as I can in 2 hours of listening to these fastidious explanations. Well made!!!
As someone who just learnt Calculus as a pre-requisite for my Calculus Based Physics course, this really helps me understand the blackmagic of 2 stepping down in x^2
solution of d(1/x)/d(x) : (as x approaches 0 ) here the lenght and width are x and 1/x respectivelty . we increased the length if the rectangle by a smol amount dx , therefore our new length us x+dx our length was represented by the variable x which is now x+dx , as our width was represented by 1/x , we can substitute the value of x there as well , so our width is 1/x+dx . the derivative of d(1/x)/d(x ) is the change in the value of the width of the rectangle (i.e 1/x) due to a SMOL change in the value of its length (i.e x) change in value of x (length) = increased lenth - initial length = (x+dx) - (x) this represent d(x) change in the value of width = decreased width - initial width = (1/x-dx - 1/x) this represents d(1/x) so d(1/x) (1/x+dx) - (1/x) = x - (x+dx)/x(x+dx) [by taking LCM] --------- = ------------------------- ------------------------------- d (x) (x+dx) - x x+dx-x on calculating we get : -dx --------------------------- [we get - dx in numerator bcoz there was a x(x+dx)(dx) - outside the bracket : - (x+dx) & we get (dx)x(x+dx) bcoz the fraction was in the form : a/b which is equal to -------- c/d ad/bc so u get the idea ig , ALSO the dx in the numerator gets cancelled by the dx n the denominator and we get: 1/x(x+dx) ] as dx approaches 0 we can substitute the value of all dx in the equation with 0 we finally get : - 1 which is also equal to - x^ (-2) [here the symbol " ^ "mean raised ------- to the power] x^(2)
How is this kind of content free?! Respect man. Seriously.
@@ricardoz5714 Yeah, I am a 13 year old. I have however whitelisted him from adblock as a small thank-you.
I also thought so. It is just unbelievable this quality is free. If this can be free, all the non-free lectures all over the world would be kind of scam.
@@pragadeeshsv6596 Thanks! I just have an interest in math.
Are you Indian?
@@justanotherguy469 Thanks!
Dear first time calculus learners, Do NOT expect to understand calculus after one pass through this video series. You must "pause and ponder" a lot, draw pictures, and see what new formulas you can discover through geometry. Read your textbook, listen to lecture, and do your homework problems, and make sure to give the sections in this video a pass or two or three (or twenty). Calculus is amazing and wonderfully intuitive, but was not invented in an afternoon, and there's a reason that the course is two semesters. If you fully understand these videos and can do computations and solve word problems, it is safe to say you have mastery over the material. Good luck and enjoy learning this beautiful subject!
Like he said, math is not an viewing sport
SybaPhoenix Gaming Online IQ tests don't count. ;)
Heck, my Algebra II class could be 1 semster...
THis would be a very good place to insert a rick and morty copypasta
+SybaPhoenix Gaming
r/iamverysmart
Someone could learn the basics of calculus from this video, but realistically, I doubt anyone could apply it without undergoing a more rigorous program.
This is more of a supplement; something to make the subject click.
Also, I've gotten an IQ of 150 on an online IQ test from choosing random answers. Online IQ tests don't count, as Danny Pike said.
Proof: ua-cam.com/video/-r2n_mK9agY/v-deo.html
We live in an age where a highly motivated individual (with internet access and time) could learn just about anything with no formal education. I hope this playlist stays available for a long time because it clarified so many things I wondered about and couldn’t articulate.
Agreed. But I will say that formal education will continue to be a thing for quite some time since some people learn better with structure and an enforced routine.
When I spent 2yrs teaching myself AI and coding, the hardest part of it all was keeping myself from going down all sorts of rabbit holes and various tangents to the point of not maintaining a central focus (as well as different sources perhaps speaking a slightly different "language" to describe things). Knowledge is so vast that no one can become an expert in it all, so reigns are needed in some form. What formal education allows for is teachers/professors having crafted a cohesive curriculum for each course, and academic counselors having aided a student in crafting a development plan to have that student's series of courses cohesively build towards a particular intellectual goal. I think what the future can bring, given this vast access to knowledge, is for people to get wiser as to how to map out a plan for themselves to most efficiently learn what it is they need to learn...since otherwise one runs the risk of learning a bunch of separate things but not particularly bringing it all together into a meaningful big picture.
Videos like this are outstanding for piecemeal insight and learning, but one essentially has to be the master of their own ship in regards to how they'll want to apply that knowledge. It's the only guiding light towards them knowing what detail to dive into next. And like you said, that takes someone motivated - it takes quite a bit of discipline.
@@chrisjfox8715to me, the problem doesn't seem to be a lack of structure in the material of this lecture series but the lack of a clear method of internalizing and remembering the concepts (like schoolwork). Unless someone watching the video is taking a calculus course, there is no clear way for them to demonstrate and retain their new knowledge and reasoning skills (besides the occasional guided question). What this series does well is it teaches concepts in a way that is very clear, interesting, and motivating. The only thing it's missing (and what, in my opinion, many stem UA-cam channels are missing) is a way for you to remember what you learned. Give this series about 100 challenging, meaningful questions that connect to some curriculum, offer some incentive for people to complete them, and this series would be pretty close to perfect for someone to teach or reteach themselves introductory calculus.
@@kaboomgaming4255
Honestly, I (paradoxically) agree with _both_ statements, as someone who learned enough for a Calc I credit over about a year, but _still_ doesn't feel right now taking a Calc II class a year later.
Hi, I'm having a problem getting -x^-2. Can you please help me out?
Area remains constant as we change the dimensions of the rectangle.
Hence, initial area = final area.
1= [ x + dx ] [ (1/x) - d(1/x) ]
1= x (1/x) - x d(1/x) + (1/x) dx - dx d(1/x)
x and 1/x multiply to give one.
We subtract one on both sides to get zero on one side.
We multiply both sides with -1.
0 =x d(1/x) - (1/x) dx + dx d(1/x)
x d(1/x) + dx d(1/x) =(1/x) dx
x + dx = (1/x) [ dx / d(1/x) ]
x^2 + x dx = dx / d(1/x)
As dx tends to 0, x dx also tends to 0, hence we can sort of ignore it.
x^2 = dx / d(1/x)
d(1/x) / dx = x^(-2)
**TIME-STAMPS TABLE**
0:06 Initial quotation
0:15 How to compute derivatives?
0:30 Why is such computation important?
0:45 It is abundant in real world
1:15 Important to always remember the fundamental definition of derivatives
1:45 D x^2 example
2:00 Graph analysis
2:40 Graphical intuition for Dx^2
3:30 dx^2 is negligibly tiny
4:20 Algebra passage to obtain derivative formula for x^2
4:45 D x^3 example
5:10 Delta volume of a cube
5:40 Negligibable parts
6:50 Pattern for Dx^a = a*x^(a-1)
7:30 Usually just symbols, but why?
8:10 We can ignore much of the terms in the computation
8:40 General case of x^2 and x^3
9:50 The importance of remembering the why
10:10 Example D 1/x
10:20 (You could just use the power rule)
10:45 Geometrical interpretation
12:00 Exercise for the viewer
12:30 Now figure out D sqrt(x)
12:40 Trigonometric functions
12:50 Geometrical view of trig functions
13:35 Starting by looking at the graph
14:10 D sinx should be cosine based on valleys and peaks, but why exactly?
15:30 Demonstration based on similar triangles
16:50 Now what is D cosx?
You're welcome
why would you do that
+Francisco Abusleme Because it adds value to the video and I think these videos deserve it, I did it also for the past video.
Ok, I just don't think it's useful
+Francisco Abusleme Also it takes about 30 minutes for me to make them, and they can potentially benefit more than 10.000 people (if 100.000 people watch the video and 1/10 needs them), so why not?
I don’t know words to express how grateful I am for 3b1b and Khan academy
I know right
Me too! Apparently 3b1b worked for khan in the past
Forgot organic chemistry tutor on that list
and crash course
he did the multivariable calculus vids in khan i think
Oh my god. When I first saw this video at the start of college in my engineering course, I didn't have any clue how to solve the 1/x and the sqrt of x derivatives via geometric analogies. Now that I quit my engineering course and am pursuing a computer science degree, I finally solved it after 5 years. I finally figured out the tricks needed to solve both equations once I got comfortable with the concepts behind calculus. It was a roundabout journey for me.
I know no one will read this, but I just wanted to share. It's a happy moment for me! Thank you 3b1b for this series.
A Small Doubt I derived them as told in the video.
The Area Lost at Top = d(1/x)*x. -> d(1/x) Change in Height, x the previous Width
The Area gained at Right = d(x)(1/x - d(1/x)) -> d(x) The Change in Width, (1/x - d(1/x)) -> The new Height
And Intutively the Area Gained = Area Lost
d(1/x)*x = d(x)(1/x - d(1/x))
d(1/x)/dx = (1/x - d(1/x))/x
= 1/x^2 - d(1/x)/x We can leave out d(1/x)/x as this will a infinitisemally small
= x^-2 ( But According to power rule it should be -x^-2)
Can you please help on it
@@muthuraj3010 remember that if the area lost is the same as the area gained you can just work on one of them. Remember that you are always working to find the slope which is (y2-y1/x2-x1). let's call the difference in the areas h, and let's call y f(x). now you have that lim (h->0) of (f(x+h)-f(x))/h = f'(x). Plugging in the values you have ((1/x+h)-1/x)/h=f'(x), now apply the common denominator to get rid of the discontinuity in the denominator, (x-(x+h)/x(x+h))/h, now simplify and you are at -h/hx(x+h) --> -1/x(x+h) Finally apply the limit (h --> 0) so that you have -1/x^2 or -(1/x^2) which is equal to -(x^-2)=-x^-2
@@muthuraj3010 I don't know if you've worked it out already. But, in the first equation of "Area Lost" the d(1/x) should be negative as it is a decrease in the area. So, In my opinion the equation should be:
-d(1/x)*x = d(x) (1/x - d(1/x))
@@Alessio216 what about proving it geometrically though
@@muthuraj3010if we were to use use new length and breadth to calculate the area? Knowing that area is 1
I keep getting these "OOOOOOH, I See, so that's why!!!" moments while watching this video. this is great.
Indeed, with this the power rule almost seems completely obvious.
If I had something like this available to me in college I would have had an A in calc based physics instead of a C-
Last video when he simplified ds/dt (t)³ to 3t², I was amazed. The power rule just from scratch
this is actually how math should be teached, because for people who are interested in the topic this is just much easier to memorize.
and i personally believe that people who are not interested in learning things like calculus should not be forced to, because it really helps noone if they are.
Math used to come really naturally to me, and after taking two years off from calculus on accident, I was struggling so hard and wondering how I even learned it the first time. I have been having those same moments like you mentioned and actually feel like I can do it!
So many educational videos on UA-cam are "edutainment" designed to give the illusion of learning something new, without actually teaching anything.
This channel bucks that trend and I am SO grateful for it. Please never stop (or at least keep going for a really long time).
Personally, I hope you eventually get into the math behind some concrete practical applications like machine learning algorithms, but I'm loving these pure math series too.
Luc Gendrot relearning calculus to get back into machine learning too👌
^^ any resources? I really want to make a (very basic, at least) neural network but I'm not sure where to start apart from 3Blue1Brown.
If you want to code one, a good starting point is this link: iamtrask.github.io/2015/07/12/basic-python-network/
I found it very useful in applying the theory of 3Blue1Brown to a real neural network. Happy coding!
This is entertaining
Luc Gendrot he has created those!!
the derivative graph of sinθ is literally mind blowing. two years of calc and it finally makes sense. thank u for giving me hope for my ap exam in a couple days, this content is incredible.
I remember taking the AP test. I got an A on the final test our teacher gave, but then I got the AP test and failed miserably. The AP test was much harder than anything my teacher threw at me.
@@alexandertownsend3291 how do you suggest to prepare?
@@singlemuskeeter6916 Study thoroughly. I didn't study enough.
@@singlemuskeeter6916 I am going to give something close to what you may call ap in you country. From most of the teacher's suggestion, its suggested trying an alternate solution for every question. And then pinpointing why did or why didn't that perspective work
@@crimsnblade8555 I'm learning calc right now. Can you explain why the two triangles are similar for the sin graph?
Our math teacher shows your videos in class!
Tushar Sadhwani he's a smart man.
You're lucky!
I just learned the formulas, and it took me a lot of time to figure out everything he's explaining in those videos
you serious? In INDIA???!!!
Which college?
Kiran Rokade school.
Thank you for existing
+
I was really hoping your channel was real.
Hola Vsauce, Miguel aqui.
I am a head of mathematics at a school in the UK and try my absolute best to teach my students and embed this sort of level of understanding. The one tool I just wish I had is animation! These animations are so clear. I use Geogebra to the best of my abilities but just can't quite offer the same visualisation as you do with these. What do you animate using? If I could just do animations a tenth as good I'd be happy. This level of visualisation adds that extra dimension for students to grasp a concept. I am very appreciative of your videos - once I have reached the limit to which I can explain something I show these videos in class to add that extra visual aid. So pleased to have your videos to complement my lessons.
I know this is a very late response, and I hope you're still the head of mathematics and my answer is still of use, but he uses a programming tool called manim in python.
@@fallow642 years late but thanks! Helpful to someone like me
Needless to say, the absolute _best_ math channel on UA-cam, not even close
I like numberphile more, LOL!
I see what you did there
To whom have you directed your commands to?
Common English: Who were you talking to?
*To whom have you directed your commands?
You don't need to include two prepositions, regardless of whether you're speaking with overly-formalized English.
Next up will be "Visualizing the chain rule and product rule": ua-cam.com/video/YG15m2VwSjA/v-deo.html
You’ll notice throughout this series that I encourage a more literal interpretation of terms like “dx” and “df” (aka differentials) than many other sources. I call this out and explain further in many of the videos, especially chapter 7 on limits, but given that students are often told not to take these terms too seriously, to be wary of treating them as literal variables, it’s probably worth adding another comment on the matter.
The path between treating these terms as literal nudges and a fully rigorous treatment of calculus is actually quite short, considering the loose language that seems to be involved. You just need to understand two things that are implicit in the notation “df” and “dx”.
First, the size of the nudge df is dependent on the size of dx. It is not its own free variable, and what it means depends on your current context.
Second, for any equation written in terms of df and dx, when you replace dx with an actual number (e.g. 0.01), and replace df by whatever nudge to the output is caused by that choice of dx, the equation will probably be slightly wrong, with some error between the left-hand side and right-hand side. But what it means to be using these differential terms is that that error will approach 0 as your choice for dx approach 0. This is why terms which are initially proportional to (dx)^2, and hence retain a differential term even after dividing by dx, can be safely ignored.
Even in the most rigorous proofs of derivative rules and properties, these tiny nudges show up, though often under the names "delta-x" or "h". The ideas presented here are essentially the hearts of those proofs but phrased without the surrounding formal language. I put together this series not just with calculus students in mind, but also with the hopes of pointing back to chapters here when I cover real analysis, the formal backbone of calculus, so I am motivated as much by an ultimate desire for people to understand the rigor as anyone else.
(Also, as a hint to those asking about how you know that the triangles at the end are similar, use the fact that the tangent line of a circle is perpendicular to its radius.)
*Yay calculus!*
I made time-stamps table for this video and the last one, have you ever thought about adding them to your videos? They increase the value very much by making them also consultable rather than only watchable (people can come back to find a particular part)
3Blue1Brown Your work is great. I refer you to all the students I tutor.
Will this series primarily be based on Calculus 1 material or will the later videos cover Calc 2 and 3 as well? Fingers crossed for some awesome multivariable calc videos.
Or you could just pin to the top (heart button) the timestamps tables I make, starting from the one in the previous video
For the f(x) = √x case, the reason why the new area is represented by dx and not df (as in the x^2 and x^3 examples) is because we square both terms in f(x) = √x to get (f(x))^2 = x. The blue area is therefore f(x) * f(x), which is simply √x * √x = x. The new area, dx, is created by a 'nudge' df(x) in both directions, which is just d√x. From there dx = 2 * √x * d√x + (d√x)^2. Ignoring the (d√x)^2 terms since they go to 0, you get d√x/dx = 1/(2√x).
I was wondering about the case and did a mistake somewhere, thanks for the explanation
Thank you. Was stuck on this for an hour.
@@novachromatic me too
wouldnt it be dx/d√x = 2√x? Or can you just switch denominator and numerator on both sides?
@@tobiasrieker1758 I guess it can be, but for the sake of this question we want to know what d√x/dx is.
Hello! I am from Brazil and would like to thank you for your work, I am a student of Industrial Chemistry and in my country we have a bad basic education, at the time there were no platforms like this, preventing access to content like yours. Thank you so much for dedicating your time to the cause of education. It is very important to many people like me.
Verdade man
Vdd
sim
Yes man eu agree let's comer a feijoada
Verdade
These videos are art... Really, they are simply works of art...
ah.. yes.. mhm of course! ..
*goes back to first video*
I have something like that in the "Maths of Relativity" series on a different channel...
@@fatitankeris6327 ScienceClic I guess?
Me too, but I’d do anything for mathematical knowledge. I’d die so that I could meet Euler and Gauss and stuff in the afterlife.
If anyone's wondering what the justification is for the claim he makes at 15:39:
The base of the small triangle is perpendicular to the right side of the large triangle. The hypotenuse of the small triangle has a slope very close to the tangent of the circle at angle theta, and therefore is roughly perpendicular to the radius shown (the hypotenuse of the large triangle). Thus, the two angles of those two triangles that are touching are about the same. We also know they are both right triangles, so that's two angles that match. There's only one possible value left for the remaining angles (sum of interior angles of triangle = 180 degrees), so all the angles match, and therefore the two triangles are similar (well, mostly, but they get more similar for smaller values of d-theta).
thank you
Thank you
how is the base perpendicular to the right side please reply I am having great mental breakdowns because of this
we need more people like you
Thanks man.
I just started learning calculus. My math teacher taught me some formulas but when I asked him "but why?" he didn't really have an answer. Until I came across this channel I had many questions. I'm really loooking forward to next chapters. Keep it up.
Show this to your teacher then! He might find it very useful for teaching.
In order to make this precise, your teacher would need more or less complex proofs involving limits.
The one time when I asked a teacher how the derivative formulas were, well.. derived, she told me to get a masters in maths. It was many years ago, and I kinda wish I had resources like these readily available back then. Maybe I wouldn't have spent over a decade avoiding everything to do with numbers, just because I was so jaded and confused.
Don’t you find the derivative using the first principle for beginners?
@Tracchofyre that's why it is important that a teacher has sufficient mastery over the subject. To teach mathematics in high school you need to have a masters in mathematics, even though you won't use 90% of what you learned in university in teaching in a high school.
But if you get too strict the supply won't meet the demand. You need a certain percentage of math, physics, etc teachers and the most talented students won't want to become a high school teacher, but they are exactly the people who can provide awnsers beyond an 'it is so because it is so' level.
Watching this series has really made me wish 16yr old me was as motivated and appreciative then as I am now at 33 of how interconnected the various maths are. I literally had a flash back to highschool and had a legitimate "ahaa" moment. This is truly excellent content!
I'm 17 and I have a great maths teacher but we haven't done derivatives yet but I'm writing a paper on the Fourier transform and I get lost very quickly and this has helped me so much with what I'm writing. Seriously, he makes great content, even with little to no knowledge about semi-advandced calculus, i understand all of it and its so great!!!
I am 17 too who gets stuck in basic maths, this video helps me to think beyond my bookish knowledge
I'm currently 16, about to enter senior high, being afraid of failing on anything at school so I chose to fail now while I have a chance
I am currently 16 and reading ur comment made me feel so grateful as this topic is going on rn in my school and this video is really helpful
Im sixteen and just want to say that i very grateful to be sixteen and can comment on your comment ( although im not a native speaker so that simple comment have 300 erros. Ps : salve do brazil a educaçao aqui é uma merda. )
I took calculus almost 6 years ago now. I'm now a grad student in robotics and diffeq is life. I love seeing how some of these things come about that either: were never explained to me or had been forgotten due to the years of plugging away.
not lifenonerx
Wait, so robotics is intrinsically tied to Differential Equations? That sounds very interesting to me and it's the first I'm hearing of it.
Hemanth Kumar I think most engineering fields use calculus, and robotics would be a category of mechanical engineering. So robotics SHOULD require it too. (Don't quote me on that. I'm not in robotics.)
It's not really just robotics. It's more like anything that is related to movement (change of position) requires the use of differential equations. Differential equations are used in most engineering fields and economics. Essentially anything to do with a rate of change can be represented by a differential equation.
Im a mechanical engineer freshmen and im planning to specialize on robotics, how was robotics? 9can i have some piece of advice? I'll appreciate it
I'm a math major currently finishing up my second semester of Advanced (i.e. proof-based) Calculus. I just learned more about why D sin(x) is cos(x) than in all my years of math up to now.
Reuben French
What is non-proof based calculus? O_o
Taxtro I'm pretty sure he was emphasizing the rigour in advanced calculus. Which is way more rigorous than calculus at the high school level...
Reuben French as a math major, what do you think about disregarding dx raised to a power? Imo it is not rigorous and well defined to just disregard a dx if we're considering derivatives in this way..
@@sergioh5515You factor dx out of everything and can then divide by dx. You then evaluate it at the limit as dx approaches 0, so anything with a dx left (i.e. initially had dx to a higher power than 1) is multiplied by zero.
Where are you now- by alan walker
For people wondering how d (cos θ) = - sin θ
Note:
While moving around the circle, sin θ is increasing but cos θ decreases from 1 to 0 and then continues its simple harmonic motion. Just use that line of reasoning and you can see at 16:56 that derivative of cos θ is - sin θ.
got it! thanks
Is what is meant by 90° out of phase. On the x axis if cosine is 1 then sine is 0 because it measures perpendicularly, i.e. the y axis direction
I wondering if for others trigonometric function is possible to find derivative from a similar way
I understand that the cosine decreases as ∅ increases. But this is only true for the first 2 quadrants. What stops us from making the same analysis on the last 2 quadrants and finding a relationship where cosine increases with ∅. The geometry of the problem would be the same. This would mean d(cos(∅))/d∅ = sin(∅) wouldn't it?
Here is my solution to 12:21
Area gained + Area lost = 0
Area gained = (1/x - d(1/x))*dx
Area lost = x*d(1/x)
Adding the areas
x*d(1/x) + (1/x - d(1/x))*dx = 0
"Distribute" the dx
x*d(1/x) + (1/x)*dx - d(1/x)*dx = 0
Rearrange to factor out d(1/x) in next step
x*d(1/x) - d(1/x)*dx + (1/x)*dx = 0
Factor out d(1/x)
d(1/x)*(x - dx) + (1/x)dx = 0
Subtract (1/x)dx from both sides
d(1/x)*(x - dx) = -(1/x)*dx
Divide both sides by (x - dx) AND dx
d(1/x)/dx = -(1/x)/(x - dx)
Distribute the terms in the denominator on the right hand side
d(1/x)/dx = -(1/(x^2 - x*dx)
The second term in the denominator on the right hand side
will go to zero as dx goes to zero.
The solution is:
d(1/x)/dx = -(1/(x^2))
Why Area gained + Area lost = 0?
I understand that it's visually correct but how we can prove this?
Can you explain why x*dx goes to zero (the last step)? I understand the whole point is that dx goes to zero but couldnt we just do it right at the beginning? Thanks a lot!
dx*dx is negligible , in reality when dx-> turns to zero derivative is calculated 18:42 in video
try this ..simple
x*d(1/x) + (1/x)dx=0
(1/x)dx=-x*d(1/x)
hence, d(1/x)/dx=-1/x^2
I think he wanted people to reason about it geometrically. Same goes for the root. Am I wrong? Then again there's no way to write a geometric solution in the comments)
That explanation of the derivative of sin at the end is mind blowing. Thank you for making these video, they're so well produced and written.
For the case f(x) = 1/x:
The blue area + red area (area lost) = 1
The blue area + green area (area gained) = 1
This implies the red area (area lost) = green area (area gained)
Red area = -d(1/x) * x
Green area = [(1/x) - (-d(1/x))] * dx = [(1/x) + d(1/x)] * dx
Since red area = green area, we have:
-d(1/x) * x = [(1/x) + d(1/x)] * dx
Dividing both sides by x * dx, we get:
-d(1/x)/dx = [(1/x) + d(1/x)] / x
Ignoring d(1/x) on the right side since it approaches 0, we have:
-d(1/x)/dx = (1/x) / x
-d(1/x)/dx = 1/(x^2)
Dividing both sides by -1, we get:
d(1/x)/dx = -1/(x^2)
Therefore, the derivative of 1/x is -1/(x^2). Power rule d/dx(x^n) = n*x^(n-1) works even when n = -1.
Another solution is: 1 = [ x + d(x) ] [ 1/x + d(1/x) ]
very nice, but I don't understand why d(1/x) should be the one approaching 0, weren't we seeing what would happen as dx->0 and looking at d(1/x)? Because otherwise d(1/x) won't get cancelled from the right side.
@@farhansadik5423 As dx → 0, the term d(1/x) on the right side becomes very small compared to the other terms.
@@eriksolis6176 I have a question.
Why is it 1 = [ x + d(x) ] [ 1/x + d(1/x) ] and not 1 = [ x + d(x) ] [ 1/x - d(1/x) ] (due to the fact that the 1/x-d(1/x) side of the rectangle is getting smaller )?
thank you
@@shiluka I don 't quit understand. if we calculate the limit of x when it's aproaching 0 the lim =0
however if we calculate the lim1/x when it aprroaches 0 it will be +∞ (plus infini)
so (what i think ) if As dx → 0, the term d(1/x) on the right side becomes very big compared to the other terms. Am i correct ?
Your reasoning of the derivative of sin(x) was beautiful. One of the nicest connections I've seen.
I didn't get why the tiny triangle with hypotenuse d(theta) is similar to the triangle with hypotenuse 1...
Nathan Richan In the limit as dtheata goes to 0, the side of the small triangle on the circle will be perpendicular to the hypotenuse of the larger triangle. You can use this fact with corresponding and alternate angles to see that the internal angles of the two triangles must match. Thus they are similar
Gregory House '' Then, because the angle between both opposites sides of both triangles with respect to θ is 90°, then the other angle on the new triangle must also be φ. '' Dafuq ?? how is the opposite side to θ of the triangle making 90° with the opposite of θ of the new is supposed to mean anything ? I mean you can have a completely different triangle having this exact same property
For those who are not understanding this, just keep rewatching this video and do not give up.
Even Im going for a 4th rewatch and now it seems that im starting to appreciate its beauty!!
same, I'm watching this video for the 4th time as well, and I've watched this video 4 years ago. I CAN FINALLY FIGURE OUT WHAT HE'S TEACHING. It actually takes a little dive into calculus beforehand in order to fully understand this video and the entire series.
Slow brain
@@Willy_WankaYou've been trolled
@@isavenewspapers8890 def not a troll bro
@@Willy_Wankaonly people who think they are stupid and want to feel like they are smart undermine others, you do it to feel better about yourself, that doesn't make you any smarter.
I wish I could have watched this video 30 years ago when I was studying calculus.
I graduated in the 90's with a BS in math and did not get beyond slope is derivative and area is integration. Man this video and others by this channel bring on a strong feeling of regret from missing how much knowledge was missing below the waterline of the calculus iceberg. Do you feel me?
@@seidomike I feel you. I took calculus 1,2 and 3 in college in the 70s. Never really understood it until watching these videos
Same but 40 years. Been bugging me ever since, had to watch this to find out.
My thoughts exactly! (well, 42, not 30)
Ok
I am a Vietnamese student, I can remember lots of derivatives but never did I understand their meanings.
But only until I find out this channel, it's enlightening!
This is a very good video explaining the reasons behind the basic rules of derivatives that school rarely or never teaches. Great job!
15:50 the reason that "little angle" is equal to θ is because the hypotenuse of the small triangle is considered a straight line, and therefore it can be considered the TANGENT of the circle. Since it is the tangent, it is perpendicular to the radius of the circle, and the rest is now obvious.
Wow thanks, i was looking for the explaination
I understand why we would consider the arc line as the hypothenuse of a triangle, but still don't understand why the triangles are similar. Why is theta back here and not another random angle ?
@@goldeer7129 looking for same answer
@@goldeer7129 scroll a bit lower to cQunc's comment and go to Guillaume's reply!
OMG thank u so much, ive been pondering for hours
An "Essence of group theory" series after this one would be awesome
I like string theory more, LOL!
I think essence of statistics is a better idea. Maybe preceded by essence of probability.
Massimiliano Tron, he's a math channel not a channel of obsolete and economically unprovable quantum physics.
Group theory is limited, field and ring theory is where it's at
It needs to be extended into modern/abstract algebra
would tune in for an Essence of Abstract Algebra series for sure!
12:27 solution
Note - sqrt(x) means (root x) i.e. (x)^(1/2)
to find - (d sqrt(x)/dx)
dx=new area
dx = sum of the areas of two rectangular strips + area of small block
dx= 2 sqrt(x).d sqrt(x) + d^2 sqrt(x)
here d^2 sqrt(x) can be neglected as has power more than one
dx= 2 sqrt(x).d sqrt(x)
1/2 sqrt(x) = d sqrt(x)/dx
hence solved.
@Gaurav verma You should be looking for the quantity d√x/dx rather than dx/d√x.
dx=2√x.d√x
1/dx = 1/(2√x.d√x)
d√x/dx = 1/(2√x)
why d^2sqrt(x) can be neglected because it has power more than one ???
@@tramquangpho dx is essentially a tiny nudge so if you square it, it is going to be so small that it approaches zero has minimal effects on the area.
Why does (d sqrt(x))^2 = d^2 sqrt(x) rather than d(x), i.e. squaring the sqrt(x) instead of the d
@@cheva1 Since you are squaring a product, you essentially "distribute the exponent" so that it becomes d^2(x). Good catch. I didn't notice that at first.
37 years since calculus in college...lights go on with this simplified and better way of teaching.
I have never appreciated the beauty of derivatives up until this video...thank you so much!
He’s explained so many math concepts better than any teacher or professor that I had. I took calc 1 and 2 but never was able to fully grasp what derivatives are, how they work. This video did explain it so well.
I'm an ex-gifted kid, probably neurodiveregent, I used to like maths at school, but now as a university student feel like I'm failing everything. Watching this series helps me both objectively understand calculus better and subjectively feel less paralyzed by self-hatred. The animations are so smooth and they keep my attention from hopping over to something else. Every video has subtitles. The whole series emphasizes understanding over memorization. What I'm trying to point out is that it's not only generally awesome, it's also pretty much neurodivergent-friendly. Don't know if it was intentional or not, but thank you in any case.
💜 hang in there mate
This is all so simple yet so profound. I love rediscovering calculus through non-hostile eyes. the whole animation involving 1/x was so elegant i loved it
This is a monumental achievement. When I was young I was exactly like the author who wants to understand every equation by visualizing how it works, only to be put off by incompetent teachers and pathetic syllabus which only cram you with queer equations without ever bothering to explain them.
Now with so many maths explanation videos out there, I have to say 3Blue1Brown is the most concise and elegant one I have seen. I don't think saying the author has done a huge contribution to human civiliaztion would be an overstatement.
Before I bumped into your channel, I had almost only algebraic intuition than the visual side. In order to make the algebraic process get etched into my intuition, I imagine that I was to explain those math concepts to some family members who were conventionally deemed as ‘have no mathy brains’, such as my brother, whose highest diploma is from primary school. And the reasoning process needs to be as plain as possible so that it fits Einstein's instruction to us: ‘If you can't explain it simply, you don't understand it well enough.’
This visual math induction of yours is somewhat like a superpower to me. And thinking about it like a superpower makes me wanna learn it. So I made watching your videos part of my morning routine. The surprising result is that now I can confidently say these two things:
Math is fun.
Getting to know math is NOT that intimidating.
Thank you. Grant.
I hope you get a prize or something for what you're doing. It's incredible
I literally have never seen(heard actually) a better teacher than you. You are actually helping us students alot by making these videos. I hope something really good happens to you someday.
Now i am no more going to give any mathematical exam, but i loved watching you videos , i wish you would've present when i was in school.
I've been applying the Power Rule so many times ever since I learned about derivatives in calc, but never truly understood why the formula is the way it is. After seeing the geometric visualizations for x^2 and x^3, it makes a lot of sense now. Thank you for making these videos, seeing all these different interpretations of formulas I didn't give a second thought about is really enlightening. I look forward to the next 7 days of videos.
Oh geez I did the viewer challenge! For once I actually completed a viewer challenge! I know people are gonna think I'm dumb for finding that "breakthrough" profound, but I did a viewer challenge!
I feel my life is complete now.
Lizard Baron woohoo!
Baby steps to giant strides!
Congratulations! It always feels good to get those in any math reference material!
Yay! Same here! Took way too long, but I did.
Best tutor everrrrrrrrr
I am a Biology Olympiad participant and I needed a good comprehension of derivatives and integral for statistics, population ecology, probability and physiology topics which I accomplished with this channel's videos.
Thanks a lot.
edit: I'm Iranian and I'm aware of the lack of fluency of English and accessibility to UA-cam among Iranian students. I would be grateful if you give me the right and cooperate with me, so I can translate your tutorials and share them with my friends.
Holy balls this series. I had one math teacher in school that taught us new formulas by going over how they are actually developed and I could understand everything perfectly, but after he left to teach at a university, I never understood what I was doing. I was just remembering how to use formulas. You have no idea how useful all of this is to me... I've already taken calculus and got ~65% in it, and it's part of my University course so I get to take it again and I've watched 2 of these and I already feel like it's going to be a piece of cake. If you are still around when I get a real job out of this course, I will repay my debt to you :))
How did it go?
@@SuperYtc1 Lol, got 95% in that calculus course and the one the following year, can't get a job 👍 gotta love going back to these cringy comments from years and years ago
@@DreadJester448 Congratulations on the 95%! I also did maths at a good uni 4 years ago and still have no job. I am learning about programming now and want to get into AI, hence refreshing calculus and this course is great. Getting a job is tough especially with all the stupid employers around. :( Goodluck!
I'm about 50 yo, all my life I was afraid of math. Calculus was a nightmare for me. With this channel, I feel like I've defeated my ancient fears. Thank you
For those who are struggling with 12:15 d/dx (1/x)
If you try to solve this the following way, you will get the WRONG ANSWER
xy = 1
(x + dx)( y - dy) = xy = 1
This is because, you have ignored the fact that dy is already negative and there is no need to put another negative sign in the (y - dy) term
So the real method becomes
(x+dx)(y+dy) = 1
xy + x(dy) + y(dx) + (dx)(dy) = 1
Since xy = 1, and dx and dy are both tiny so their product will be negligible
x(dy) + y(dx) = 0
x(dy) = -y(dx)
dy/dx = -y/x
Since y = 1/x
d(1/x)/dx = -(1/x)/x
d/dx (1/x) = -1/x²
You are a genius man. Thank you!
According to the video i write it this way :
df = 1 - 1 = 0 = (x +dx)*1/x - x*(1/x-d(1/x))
U juste calculate and get the answer
Stay geometric and u look d(1/x) as being a distance (so positive) and use - signe when needed
For the square root it's even simplier : just use the same drawing as for x square (replacing both x on x and y axes by √x on both) and u get the answer as (x)' = 1
correct me if I am wrong, "x*d(1/x)+1/x*dx=0", right? "x*d(1/x)" is the area lost and "1/x*dx" is the area gained but the area lost and the area gained are the same, so we put "0" on the right side.
wow this good stuff
How do you get from this expression:
x * (dy) = -y * (dx)
this expression:
dy / dx = -y / x
?
It is equivalent as if from this expression:
x * a = -y * b
you get this expression:
a / b = -y / x
right?
How did you derived it?
15:39 why the triangles are similar (commenting so i can look back at this, except im not a big brain math genius like everyone else here)
- big triangle angles: θ + 90°+ (other angle)= 180°, so θ + (other angle) = 90°
- radius/big t's hypotenuse is perpendicular to tangent line of circle (hypotenuse of small triangle)
- knowing that alternate interior angles are congruent, angle btwn radius and bottom part of small t is θ
- because angle btwn tangent line and radius is 90° (hypotenuse of small and big triangle), 90°- θ = (other angle)
- this means that the far right angle of small t is "(other angle)"
- because small t has a right angle and has (other angle), and θ + (other angle) = 90°, the last angle is θ.
- because the angles of both triangles are the same, they're similar
Thank you so much! 👍🙇♂️
you could just have made sure that the triangle you drew had the same theta of your original triangle "theta", as you can do it for any d(theta)
Thanks ! As a person who hvn’t touched geometry for years, I understand the proof instantly
you are math god. It took me years of study and even more research to understand the essence of math. Wish u existed 10 years ago :(. I had only one good math teacher in collage, but u outshine everyone. Your explanations are simply beautyful, intuitive and simple. When i was studying i had the same approach to math problems. PLEASE PLEASE continue your work. I would like to see you explain FUNDAMENTAL FORMS, FOURIER SERIES AND SPHERICAL HARMONICS. I had very hard time to understand those. I consulted countless professors and used Bronstein math manual, wolfram wiki, everything. Still those are still abstract subjects to me. Pls help
Taking a step back to remember why the power rule works is literally why I'm watching this series, so thank you!
I am fairly sure he is the greatest mathematician of our time. His ability to find the deeper truth behind common maths is simply brilliant.
More like greatest teacher
I am thirteen and now I'm here listening to 3B1B talk about the essence of calculus. He truly simplified the complex and 'boring' ideas into some simple but beautiful, wonderful, and interesting drawings. Best respect.
What I think is that instead of turning boring things into interesting things, he shows how beautiful maths is, as what it actually is. Maths is always attracting me, like how bread attracts a starving person.
@@candiceyang7956 lmaooo love the analogy at the end
Damn, 13yr olds are from 2010. I feel so old now!
15:23 when he switched voices it kinda scared me loll
Me too :)
haha me too!!
3:58 Maybe a better explanation than "this is so tiny, you can ignore it (nevermind the other term also gets truly tiny and will not be ignored)" would be to actually divide by dx once to get df/dx=2x+dx and let dx approach zero, so that df/dx approaches 2x.
EDIT: You did actually explain it this way at 6:11 :)
Excraciation because in math is it prohibited to say ignore these terms. Math is an abstract science, not like the physics
12:22 Area gained + Area lost = 0. Area gained = dx*1/x. Area lost = x*d(1/x). --> dx/x + xd(1/x) = 0 --> d(1/x)/dx = -1/x^2.
Under-rated comment right here.
This really helped clarify, but could you help explain why area gained shouldn't equal area lost? Wouldn't this necessarily mean then that Area gained - area lost = 0, meaning that the total area remains at 1?
Subtle tweak to make (6 months late).
Area gained is actually (1/x - d(1/x))*dx. Expansion gives you a dx*d(1/x) term. When you divide by dx, you are still left with a d(1/x) term which will still be really small, so we just forget about it (as with dx terms in previous examples). The answer ends up being the same, but this way clarifies the similarity between this and the x^2 case :)
I dont get why area gained+ area lost =0 !!!!!
Đạt Trần Because we're defining the box to have an area of 1. When we add on another section, we know that the area removed must be the same so that the area of the box stays the same.
for those wondering about the derivative of 1/x here's what I did :
the text stated that the area lost must be equal to the area gained in terms of value so theu must be of opposite signs :
x *d(1/x) = - (1/x)dx
>> d(1/x) = -(1/x^2)dx
>> df/dx = - 1/x^2 (where f = 1/x)
hope this is a logical reasoning
I’d love to say a huge thank you to you. All the videos you have made are absolutely fascinating and beautiful. I remember being so deeply moved by maths when I first saw your topology videos. They have motivated me a lot to pursue mathematics in my further studies and I am so glad to have you to be my best maths teacher. Don’t stop making videos and thank you very very much!!!
Here's what I got for the f(x) = 1/x problem.
Looking at the small rectangle with sides dx and d(1/x), we know that the derivative is the ratio of its height over width (its slope) as dx approaches zero.
Using the graph, we find the width = x + dx - x = dx, and (remember to substitute in x + dx) the height = 1/(x + dx) - 1/x = (x - x - dx)/(x*[x + dx]) = -dx/(x^2 + x*dx).
So then the slope = (-dx/[x^2 + x*dx])/dx = -1/(x^2 + x*dx).
Now as dx shrinks to 0, so does the x*dx term in the denominator, and we are left with -1/(x^2).
thank you for your explanation! I finally understood that problem because of this
Here's an alternative solution:
We know that the area lost equals the area gained, so we can make the equation:
-x * d(1/x) = dx * 1/x
^ (important note: the left side, being removed, is negative. this is just a rearrangement of x * d(1/x) + dx * 1/x = 0)
Now, remember that our goal is to find an equation that will leave us with: d(1/x)/dx = some value.
Let's try rearranging the above equation to get to that point:
-x * d(1/x) = dx * 1/x
1. divide both sides by -x - note that this is *not* addition so you don't need to distribute the division. You may think of it as multiplying both sides by 1/-x
= d(1/x) = dx * (1/x) * 1/-x
2. divide both sides by dx
= d(1/x)/dx = (1/x) * (1/-x)
3. simplify
= d(1/x)/dx = -1/x^2
4. we've found the answer!
@@NateLevin "-x * d(1/x) = dx * 1/x"
Why is only x negative here? How is it negative?
"note that this is not addition so you don't need to distribute the division."
What does this mean?
@@qleo1769 If you look at the red and green part, the area lost is equal to the area gain, the loss of area represents that negative notation at the left side of this equation
I tried to get it using the nice equation in the first episodes. A small change in y divided by a small change in x. I got (sqrt(x+dx)-sqrtx)dx. I tried thinking of ways to simplify it, and nothing looked like ti would work until I set it equal to another variable, which I called y. I got x + dx = dx^2 y^2 +2dxysqrtx +x. Miraculously, the x cancelled out and left me with dx=(dx)^2*y^2+2dxysqrtx, which is incredibly easy to solve, and so left me with -1/(x^2)
In the czech language, the word education comes from farming, where farmers would "educate" the land - prepare it for actually growing (in work, but also life in general). I think this series is exactly what education is supposed to be. Thank you!
Grant, Australian viewer here
As an aspiring physicist and mathematician, I’ve always loved these vids. Recently I was invited to participate in our schools year 10 accelerated maths course (currently year 9) and so to prepare I’ve been self-teaching some introductory calculus. These videos are just sublime, I’m learning so much and can now almost consistently replicate your results, always using the visual interpretations. The videos are legitimately going to change my life and career, so thank you so much for this amazing content!!
This video was sooooo satisfying to watch, especially considering how when I was learning calculus, I'd ask, "Wait, why does this differentiation rule work?" and my teacher would say, "Oh, it's just derived from the limit definition of a derivative."
mollymack98
Well, it is.
None of those rules are defined by pretty pictures, but visualization helps us to build a mental heueristic, which enables us to find proofs.
The main thing to keep in mind is that the limit definition did not come before the visual understanding by the mathematicians who invented it.
A better name for this channel - pause, ponder and rewatch!
Why?
@@facehugger4145 because that's what your supposed to do
my god how can anything be this good
How can anyone be this good?
For the challenge at 12:27, I propose the following solution:
d(x) is the new area (i.e the yellow area)
That mean we got:
d(x) = d(√x) √x + d(√x) √x + (d(√x))²
Which we can bring to:
d(x) = d(√x) (√x + √x) + (d(√x))²
If we divide both sides by d(√x):
d(x)/d(√x) = 2√x + d(√x)
If we take the inverse of both sides we get:
d(√x)/d(x) = 1/(2√x + d(√x))
And as d(√x) tends to zero it becomes negligible and we finally get:
d(√x)/d(x) = 1/2√x
Which is the derivative of √x, hope that helps.
Thank you for your comment! You helped me connect the dots backwards.
I guess I was biased by how the input “x” was “length” in previous examples, but here it’s area.
The challenge I faced was in verbalizing the function to aid its visualization…
x^2: gives area of square of length x;
1/x: gives height of rectangle of length x such that area is 1;
x^(1/2): gives length of square of AREA x;
@@vaguebrownfox I'm French so I didn't understand some of ur words but I'm glad it helped u.
Help me understand, why did we take the inverse? I understand it's to satisfy the "d√x/dx" part, but why does that have to be the form of the answer in the first place?
@@cendolgbf remember, each time u calculate the derivative u gotta divide the tiny vertical nudge which is d√x in this case by the tiny horizontal nudge which is dx in every case, moreover if u look throughout the video he always calculates the derivative of a function f with df/dx, so remember the tiny horizontal nudge which is dx is always in the denominator, that's why I took the inverse, it's not just cuz it's a useful trick I pulled out of nowhere, hope that helps.
@@azizautop995 right, i was confused by the subtle change that d(x) is now the area instead of the length (in other examples). That makes a lot of sense now, thanks!
I cannot stress enough how helpfull this has been. Going through Highschool and Uni where only surface level explanations are given can dissolution you and make you forget why you ever liked math and science in the first place. These videos are helping so much to re-ignite my curiosity and remind me why math and science exited me so much in the first place.
12:21
If you zoom in the point the ratio d(1/x) / dx (height of that small rectangle/ width of the small rectangle ) should be the same as the ratio of the bigger rectangle (1/x) / x hence it is 1/x^2 and ofcourse the sign is because d(1/x) is actually negative.
-d(1/x) / dx = (1/x)/x
d(1/x) / dx = -1/x^2
Maybe you guys can think like this way easier: In order to keep the area equals to 1 all the time, the new amount of gain【(1/x)*dx 】must equals the area of loss【x*d(1/x)】 when the tiny x change. This gives us the equation(1/x)*dx=(x*d(1/x), then we can get new equation d(1/x) / dx=(-1)*1/x^2【since the nominator change is negative we need to times -1】
Can you please elaborate your thoughts on 12:33 too?
@@ming5363 DISCLAIMER: assume dx in infinitely small, so an incredibly small increase of dx would yield an incredibly small decrease away from our 1/x, which would practically retain that original side length. So 1/dx would approximately be = to 1/x
Here is my best analysis
d(1/x) = df
He says df is a negative length, ok
x*df + (1/x + df)*dx = 0, after dx occurs the height of rectangle to the right is not exactly 1/x anymore it reduces by df. remember, df is a negative length in diagram above.
-x*df = dx/x + df*dx
df*dx is negligible and essentially zero, cancel
-x*df = dx/x
df/dx = -1/x^2
@Sotobito Well no.. The area of gain is 【(1/x)*dx 】. Why are you trying to add the area of that little transparent box on top of green filled rectangle? That green filled rectangle is what represents the area of gain and its equation is 【(1/x)*dx 】only
There is magic in your videos... concepts become crystal clear
The Solution at 12:16
Since the area should remain constant that is 1unit.
Therefore area of rectangle
(x+dx)*(1/x-d(1/x))=1,
x*1/x-x*d(1/x)+dx/x-dx*d(1/x)=1,
since dx*d(1/x) is very very tiny value we neglect it , therefore
x*1/x-x*d(1/x)+dx/x=1,
1-x*d(1/x)+dx/x=1,
x*d(1/x)=dx/x,
d(1/x)/dx=1/x^2.
Since, d(1/x) decreasing the height of the rectangle we take symbol to be negative.
You mean 1/x + d(1/x), not 1/x - d(1/x). We want 1/x to decrease by a tiny amount, so we should add a negative value, that being d(1/x).
Waw. Thank you so much !
Your videos are concise, entertaining, and poetic. I'd love to see the Essence of Probability series!!
This is so good. My second time watching and this time taking notes and drawing some of the diagrams. I am so grateful for you sharing your experience.
Man the nostalgia. I watched this a couple years ago and I still think it's the best introduction to calculus ever
Your videos are so beautiful. They really express the pure beauty and elegance of mathematics and also physical phenomena. Unfortunately, not everyone on the earth can or wants to experience this beauty. I am privileged. Thanks !
Have you thought about doing more videos over complex analysis?
The explaination by 3b1b is amazing but lets pause for a moment and appreciate the animation that helps us visualise all the concepts so clearly
Hi, I'm having a problem getting -x^-2. Can you please help me out?
Area remains constant as we change the dimensions of the rectangle.
Hence, initial area = final area.
1= [ x + dx ] [ (1/x) - d(1/x) ]
1= x (1/x) - x d(1/x) + (1/x) dx - dx d(1/x)
x and 1/x multiply to give one.
We subtract one on both sides to get zero on one side.
We multiply both sides with -1.
0 =x d(1/x) - (1/x) dx + dx d(1/x)
x d(1/x) + dx d(1/x) =(1/x) dx
x + dx = (1/x) [ dx / d(1/x) ]
x^2 + x dx = dx / d(1/x)
As dx tends to 0, x dx also tends to 0, hence we can sort of ignore it.
x^2 = dx / d(1/x)
d(1/x) / dx = x^(-2)
@@akankshagupta4138 i cant see any mistakes but you have done a great job, i was looking for this as i am stuck on this myself
I'm 14 years old and learned how to do calculus only by watching this videos and practising with example tasks/playing around with the thoughts. Great series, great channel, great explained animated math! I suggested it to all my friends. Thank you 3blue1brown!
These videos should be shown in schools not the pile of formulas.
It’s imperative that 3B1B is kept on the internet forever, for everyone. The value of this channel cannot be quantified. I’ve learned so much from Grant.
I'm on a rewatch of this series, and wow! This episode is still mind-blowing
Fantastic. The best illustration ever. ❤
you know what , i dont want to write a lot in comment , just 3 works i will dedicate to you.... YOU ARE AMAZING
LOVE FROM INDIA
My brain has to work so hard to wrap around this stuff, but when it finally does its so so satisfying.
Key to find the solution is: searching for a conncection between dx and d(f(x)) and if necessary cancel out the parts, which get smaller and smaller for a tiny dx:
Solution for 11:31:
The derivative of the function f(x) = 1/x is to be sarched:
The visulation shows the input as the length of the rectangle and the output as its hight, while the area stays 1.
You recognize that the area removed has to be the same as the area added:
Area removed: x*-d(1/x)
Area added dx * (1/x)
x*-d(1/x) = dx * (1/x) /// divide dx, divide x
- d(1/x) / dx = (1/x) * (1/x)
d(1/x) / dx = -(1/x) * (1/x) = -x to the power of -2 = the derivative of 1/x
Solution for 12:28:
The derivative of the function f(x) = sqrt(x) is to be searched:
This visulation shows the input as the area of the square and the output as its length.
dx is the added area; d(sqrt(x)) is the added length;
The added area dx can be expressed with d(sqrt(x)):
dx = 2 * d(sqrt(x))*sqrt(x) + d(d(sqrt(x)) squared
The squared edge arrives 0 for smaller and smaller dx -> can be canceld out; All we need to do is, solve the equation:
dx = 2 * d(sqrt(x))*sqrt(x) /// divide dx
1 = (2 * d(sqrt(x)) *sqrt(x)) / dx /// faktor 1/2; factor 1/sqrt(x)
0,5 * (1/sqrt(x)) = d(sqrt(x)) / dx = the derivative of sqrt(x)
Thanks, but I don't understand how we choose which variable represents the area for example in the visualisations, sure they work for these examples but I have no idea how to do them myself when sometimes dx represents the length added and other times the area..
I'm from China, and my words just fails to express my gratitude to 3B1B! I'm a student from data science, and I really needed it to refresh my memory ! Thanks a lot!
I'm only in seventh grade, and though it takes me a lot of time to understand this stuff and probably need way more learning in the areas of linear algebra, this guy does a great job helping me understand core conceptual ideas of a part of math I won't be taught until high school
Hey, Professor Bertrand! For the first homework exercise, I found the answer by adding the areas of the rectangles (the red and the green), setting that sum equal to 0 and solving for d(1/x)/dx. For the second homework exercise, I found the answer by using the same idea as with the square with side length x, replacing x with sqrt(x), and then solving for d(sqrt(x))/dx. Finally, I compared these answers to the answers that would have been expected from using the power rule, and found that they were exactly the same. Thank you so much! I really enjoyed this video! Keep up the good work!
@@Dhruvjindal0426 I can help you with that sometime.
OH MY GOD YOUR METHOD WORKED THANKS SO MUCH I HAVE BEEN STUCK ON THESE FOR HOURS
I was introduced to calculus 3 years ago and I have hated it since. I couldn't have believed it was this beautiful. I love you man!
Your videos have taught me to imagine a lot. I'm an aspiring data scientist and many of my friends follow your content. A request will be making such short series of probability and statistics series. Would be really helpful.
Watching these videos I can see how Math education in schools suck really really bad! Why the hell on earth we didn't have such methods and teachers to teach us Math. I could still use a teacher who knows math by heart and can teach even this is my junior year in University. I hope people who memorises Math and think they know Math, stop being teachers and leave some space for teachers like this.
dunno, but i would probably struggle to understand things like this when they are presented only geometrically. So i doubt i would understand what it is all about. for example, why triangles presented are similiar?
With such marvellous videos starting to come up more frequently on youtube, I can foresee an era where school education becomes redundant.
They are both rectangles, so only one more angle is needed but I cannot find it neither
The difficulty with always teaching conceptually/intuitively is often time, a packed curriculum, and students who come in with a poor understanding of pre-requisite material. I teach calc I, and focus on intuitive aspects as much as possible, but time is always against me, especially when students struggle with pre-requisite material and, as a result, struggle to see some of the beautiful underlying calculus.
Cowmoo83 Yes, you are right with the time issues. The teachers who teach pre-requisites should be better so that you could be better as well. But I think it's not a simple problem to solve. There are some pretty good approaches with education. For example Sal Khan's learn or repeat approach as I call it, is a good one. You can find it on his TED talk. I myself, believe that as humanity we are well below our potential. To reveal that potential, we need superior education for everyone.
I watched this video 2 years ago when I took calc 1, and I think it’s what really made me fall in love with this channel.
12:21
We have that xdf + dx(1/x - df)=0 since the area remains constant.
I isolate df/dx with simple algebra obtaining df/dx= -1/(x^2) - df and since df is negligibly small we can cancel it.
For the other we have that dt^2/dt is equal to 2t so i just take the reciprocal of both parts and substitute t=sqrtx so i have that the derivative is (2sqrtx)^-1.
Somehow the very first step I did was already wrong.
My initial equation is:
x * 1/x = x * 1/x + dx * (1/x - d(1/x)) - x * d(1/x)
Basically saying:
The area = the area + the stuff on the right - the stuff on the top.
I guess I have to keep d(1/x) negative even in the "stuff on the top" part.
For anybody wondering, I gave up once I had:
d(1/x) / dx = 1/(d(x) * x) + 1/(x²)
If the area remains constant, doesn't that mean that xdf = dx(1/x-df) ? I get df/dx=1/x^2 - df/x , so the opposite sign. Where am I wrong with my logic?
i got it ln x
As he said (11:55) you sould consider df as a negative quantity since the function is decreasing.
One of the best vídeo of UA-cam.
Well done! I have finished my first half semester in Calculus 1 at a private University and haven't learned as much in 2 months as I can in 2 hours of listening to these fastidious explanations. Well made!!!
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Best math channel i have come across one youtube. Never stop doing what you are doing!
As someone who just learnt Calculus as a pre-requisite for my Calculus Based Physics course, this really helps me understand the blackmagic of 2 stepping down in x^2
solution of d(1/x)/d(x) : (as x approaches 0 )
here the lenght and width are x and 1/x respectivelty .
we increased the length if the rectangle by a smol amount dx ,
therefore our new length us x+dx
our length was represented by the variable x which is now x+dx , as our width was represented by 1/x , we can substitute the value of x there as well ,
so our width is 1/x+dx .
the derivative of d(1/x)/d(x ) is the change in the value of the width of the rectangle (i.e 1/x) due to a SMOL change in the value of its length (i.e x)
change in value of x (length) = increased lenth - initial length = (x+dx) - (x)
this represent d(x)
change in the value of width = decreased width - initial width = (1/x-dx - 1/x)
this represents d(1/x)
so d(1/x) (1/x+dx) - (1/x) = x - (x+dx)/x(x+dx) [by taking LCM]
--------- = ------------------------- -------------------------------
d (x) (x+dx) - x x+dx-x
on calculating we get : -dx
--------------------------- [we get - dx in numerator bcoz there was a
x(x+dx)(dx) - outside the bracket : - (x+dx) &
we get (dx)x(x+dx) bcoz the fraction
was in the form : a/b which is equal to
--------
c/d
ad/bc so u get the idea ig , ALSO the dx
in the numerator gets cancelled by the dx
n the denominator and we get: 1/x(x+dx) ]
as dx approaches 0 we can substitute the value of all dx in the equation with 0
we finally get : - 1 which is also equal to - x^ (-2) [here the symbol " ^ "mean raised
------- to the power]
x^(2)
hey y did u watch that solution when it was ur HW 🤣🤣🤣