Higher order derivatives | Chapter 10, Essence of calculus

[@derekdziobek5998]

The change of position over time is velocity. The change of velocity over time is acceleration. The change of acceleration over time is a jerk. The change of a jerk over time is an election.

[@balaportejean7015]

when know who it is ahahhahaha

[@Fujibayashi50]

@Spaced without a trace Cool story, bro

[@cletushumphrey9163]

@Spaced without a trace at what certain point in time did anyone ask

[@johnnypiquel2295]

@@cletushumphrey9163 did anyone ask you to reply ?

[@mannyheffley9551]

@Spaced without a trace fax

[@pillsofpink2546]

My calculus professor is sending us links to these vids instead of having a zoom lecture. So congrats on teaching MATH155 at Colorado State University.

[@confusedsperm9521]

Bruhhh when free online material is bettah than paid University teaching , I love the future

[@mau345]

Ahahaha honestly though its the best for everyone

[@bernhard8051]

So you pay a huge amount of money and they don’t even bother to do anything?

[@RandomDays906]

The 4th, 5th, and 6th derivatives are Snap, Crackle, and Pop, respectively.

[@jamesmnguyen]

Dominic Boggio Lock and Drop

[@TheZenytram]

lol this is really true.

[@buxkhurana]

yo can u tell me a good source to learn this pls thanks

[@dqrksun]

@@buxkhurana Wikipedia

[@fabianc.vargas5166]

how i met yout mother reference?

[@tobybartels8426]

4:48 : I have to correct this, because it confuses my students too. You said ‘A negative second derivative [of displacement] indicates slowing down’, but that's only correct _if_ the velocity is positive. As you noted in the video on derivatives, a negative velocity means that you are headed in the negative direction. And in that case, a negative acceleration means that you are _speeding up,_ with the velocity becoming even more negative, while a _positive_ acceleration means that you are slowing down. If you want a quantity that's positive when you're speeding up and negative when you're slowing down, then you need to take the derivative of the _speed,_ that is of the absolute value of the velocity, so the second derivative of the total distance travelled, but _not_ the second derivative of the displacement. (Arguably, this fits more with the way we use the word ‘acceleration’ in ordinary language, but the technical meaning is the second derivative of displacement.) As an aside, this disparity becomes even more extreme if you're moving in multiple dimensions of space. In that case, the displacement, velocity, and acceleration are all vectors, and it doesn't make sense to say that they are positive or negative as such. Then the speed is the magnitude of the velocity vector, and the derivative of the speed is again positive if you're speeding up and negative if you're slowing down. But now it's also possible for the derivative of the speed to be zero, even if the acceleration is nonzero! In that case, the speed is constant but the velocity is not, because you're changing direction.

[@bonniejacques9176]

Came here to say just this. Thanks!!

[@tobybartels8426]

@@bonniejacques9176 : You're welcome! I really went on about it, didn't I?

[@uncleswell]

@@tobybartels8426 this is the kind of setting and content where you should go on about it. I really appreciate you taking the time to share this.. thank you.

[@tobybartels8426]

@@uncleswell : You're welcome!

[@amanpants275]

Isn't the negation of second derivative gives max of function

[@Ash-bc8vw]

Thanks

[@dannyundos8927]

I think Korean is funnier here. After "velocity", you just add "가". Displacement = 변위 Velocity = 속도 Acceleration = 가속도 Jerk = 가가속도 4th derivative = 가가가속도 5th derivative = 가가가가속도 6th derivative = 가가가가가속도 ... nth derivative = (가)^(n-1)속도

[@새싹-l9z]

nth derivative : gagagagagagagagagagagaagagagagagaggagaagagagaga.....gagagasokdo

[@doctorlove5415]

this is beautiful

[@Anonymous-jo2no]

Wait... the Korean for "velocity" is sokdo? I smell loanword here... (速度/そくど) wwwww Yes, of course I know the word in both languages is a loanword from Ancient Chinese...

[@youknowwho8925]

Exactly same as Cantonese

[@이효건-o4o]

Amazing I didn't know that

[@mesplin3]

3:47 "Interestingly, there is a notion in math called the 'exterior derivative' which treats this 'd' as having a more independent meaning, though it's less relatable to the intuitions I've introduced in this series"

[@DavidSartor0]

Thank you.

[@kjekelle96]

0:00 intro 0:39 derivative of the derivative 1:53 notation 3:58 intuition 5:05 outro

[@noone3367]

This channel deserve more subscribers

[@300483rahul]

ebulating thats great, this guy deserves millions of dollars per video:)

[@JorgetePanete]

MOHAMED DHYA KAHLAOUI deserves*

[@JorgetePanete]

Rahul Jobanputra that's*

[@kimothefungenuis]

1 M subscribers now

[@thedancingbudgie8045]

2.3 million subs now

[@xtuner88]

Who dislikes this video is a 3rd derivative

[@dijek5511]

whoever, or those who

[@EriqireM]

*ahem* whomever

[@dijek5511]

Actually whoever though, because it is a subject and not an object :P.

[@EriqireM]

Is the video the subject, or is the individual the subject? "Whomever" isn't incorrect its just impolite, which reinforces the joke.

[@patrickhodson8715]

Edward McCarthy no, it is incorrect because "whomever" is the object case. It's like saying "him went to the store" instead of "he"

[@idrisShiningTimes]

Beautiful explanation, visualisation, and most importantly, the simplicity you always use to explain complex terms. Love it

[@chaosui3169]

3:31 much clear now: the second derivative is treated as the difference of two first derivative: if its positive, it increases

[@aajjeee]

Position Velocity Acceleration Jerk snap Crackle Pop

[@AvinashtheIyerHaHaLOL]

you forgot displacement

[@MCPhssthpok]

Barnesrino Kripperino I was taught velocity, acceleration, jerk and jounce.

[@Wherrimy]

Also, Jounce (d(Jerk)/dx), Absement, Absity...

[@aajjeee]

neither jounce nor snap is accepted widely, but there is an informal rule that the higher orders are snap crackle and pop

[@swiminbandgeek]

Barnesrino Kripperino you don't have to be a stick in the mud

[@patrickhodson8715]

Nowadays everyone is releasing non-episodes in the same universe. First there was _Rogue One: a Star Wars Story,_ and now we've got _Higher Order Derivatives: a Calculus Story._

[@SandeepSingh-qr3dk]

Hello Grant, I really admire your videos as you can see I am watching these again even after two years. Please do a series of animations on Complex Analysis and Transforms (laplace, Fourier and Z).

[@SuperElephant]

-5 >> Absounce -4 >> Abserk -3 >> Abseleration -2 >>Absity -1 >>Absement 0 >> Displacement 1 >> Velocity 2 >> Acceleration 3 >> Jerk 4 >> Jounce I really had a hard time understanding Less than 0 and more than 2... Can anyone make a video to explain it all??

[@bace1000]

Absement is just displacement multiplied by time, i.e. how far an object is from a point and for how long it has been there. It is constant only if the object is not displaced, but is steadily increasing if the object is displaced.

[@oldcowbb]

and you can do a half derivatives

[@buxkhurana]

yo can u tell me a good source to learn this pls thanks

[@ThePharphis]

Is there an interesting and readable source on half derivatives? I only heard about their existence a year ago and I'm pretty curious

[@dadgumit2505]

negative derivatives are just integrals right?

[@loganstrong5426]

I took Calculus (1 2 and 3) back in high school. I am watching this series for probably the third time because these were all the same intuitions I had that helped me understand the subject the first time around. Keep up the great work with all your videos!

[@hahahasan]

You should definitely do a video on the gamma function and fractional derivatives.

[@oidazaubara]

The "change of how the function changes" really made it click there. Thank you.

[@unclegranpawafiaahmedyahia5925]

Ces vidéos sont supers..je conseil ; grand merci 3bleus 1marron..

[@robwhitlock5030]

3:17 Why is d(df) proportional to (dx)^2?

[@iabervon]

Rob Whitlock It helps to work it out for something like f(x)=x^2, like in the earlier video about the derivative of x^2. In that, df was 2 rectangles, x by dx. Now, ddf means that you add another dx to x in the df illustration, which puts a dx by dx square on each rectangle. The area of this pair of squares is 2dx^2. If you go through the example derivative illustrations, you'll find that they each work this way (cubes add 6 x by dx by dx boxes, sin has a tiny triangle on a tiny triangle, and so on).

[@艾歐小塊玻璃]

I’d like to share a example of f(x)=x^2 I think of it d(df) as the difference between the 2 df just like they were in the video. so d(df) = df2 - df1 If f(x)=x^2, df = 2•dx•x (like the 2 rectangles in the earlier video) d(df) = df2 - df1 = 2•dx•X2 - 2•dx•X1 (Just like the video, let X2 = X1 + dx) Factor the 2•dx out We get 2•dx•(X2-X1) = 2•dx•dx So, it seems like that ddf is proportional to (dx)^2 in this example

[@ryanlira7194]

can u do an essence of differential equations? ubhave no idea how much i love these

[@marcinukaszyk4698]

I just want to say:thank you! I learned a lot

[@ghostofastarman4479]

It looks like this series is going to end the day of my AP Calculus exam. Thanks for helping me study +3Bue1Brown

[@YunsuPark-xz2uu]

3:54 does anyone know why (dx)² becomes dx², not d²x²? I know everyone writes second derivative like that, but I'm just curious. Is that simply because dx² is almost same as d²x²

[@CepheusMappy]

It is the same. I heard that its because it would be messier to write d²x² instead of dx²

[@isavenewspapers8890]

d isn't a variable. It means "a tiny change in", so dx means "a tiny change in x". We treat "dx" as a single object, so dx^2 just means dx * dx.

[@tymothylim6550]

Thank you very much for this video! It was quite informative seeing how the 2nd derivative can be a comparison between two sets of 1st derivative value multiplied by some dx

[@andreranulfo-dev8607]

4:31 Just wow! Now I trully understand inflextion point!

[@alyssabowen9297]

Oh my gosh, thank you. I finally understand now. I was having a hard time figuring out the relationship between f(x), f'(x), and f''(x) but the displacement, velocity, and acceleration explanation made so much sense.

[@severussnape5171]

you must be some kind of god...thanks for these awesomely illustrated and explained videos Sir!

[@topilinkala7651]

For best understanding why the derivative of accleration is called jerk imagine a computer driven lathe. To move the tool to position you want smooth movement so that the tool does not break. If your movements jerk is too much then the movement is not smooth but it's jerky. Another example of jerk is in an amusement park. If you ride the coffe cups the movement of those cups have sudden jerks in them and if you graph the movement function and calculate jerk you find out that jerk is high on those parts of the movement. So the name jerk is a very good description what changing acceleration means. Btw. Human's sensory system work well in acceleration and so smooth acceleration does not cause any feelings in itself. For example your inner ear does not react to gravity. A non changing acceleration field does not register. But increase jerk and you inner ear starts to function. That's why amussement park rides use high jerk to cause effect in humans.

[@existentialchaos8]

When I was around 9, I realized that all number patterns have "layers" underneath them. The first layer below it would be how much it increased by each time, the 2nd would be how much the 1st layer increased by each time, and so on. I had this theory that every pattern, if you "peel" the layers enough, it would always reach a layer where all terms would be the same number, and that was the "base layer" that every pattern was made out of (now I know this is true for polynomials functions), and each pattern could be classified by the number of layers it had. For example, for a pattern like 1, 4, 9, 16, etc., it would be a 3rd layer pattern because the layer underneath, or the 2nd layer, is 3, 5, 7, 9, ..., and the layer underneath that, or the 1st layer, is just 2, 2, 2, ... I realized I just basically found out the concepts of arithmetic sequences, polynomial degrees, derivatives, and possible Taylor Series.

[@hugoandre96]

thank you very much, I have been using your series on calculus to help me study for my final. you have helped me better understand some things I didn't understand in class, such as how limits and implicit differentiation

[@vigneshbalaji21]

Awesome explanation of order of derivatives. Intuitively explaining rate of change of slope as second derivative.

[@vitoriaxavier4233]

tenho vontade de chora de tanto q amo esse canal it means i love this videos so much that i wanna cry

[@ErmisSouldatos]

this is so well explained and intuitive. why can't all teachers teach it this way instead of boring formulas and telling you to stfu when you ask why this is so, which is what my teacher did all the time? Did he have to be such a d^3s/dx^3 ?

[@loganborghi5727]

the double upload made my day, thanks

[@freddyfozzyfilms2688]

All hail our great leader 3b1b.

[@linazso]

this notation was really strange for me, so thanks for clearing that! :)

[@AJ-er9my]

Excited for the main event! Thanks for explaining this

[@feicuitadie]

3b why no quote at the beginning of this video? I love all those quotes you had in other videos

[@ahmedgaafar5369]

incredibly amazing as usual.

[@YourMJK]

4:10 Does anyone know which type of function that is?

[@YourMJK]

I think I got it: it's roughly -1/2 * (sin(x)-x)

[@jackfelag6923]

If you want the official term for it, it's called a sigmoid. That's assuming it approaches horizontal asymptotes when x approaches either negative or positive infinity. Some functions that are like it are hyperbolic tangent (tanh(x)), inverse tangent (arctan(x)), and 1/(1+e^-x)

[@cameronadams4366]

-cos(x) + 1? But the derivative of that (which is the velocity) is sin(x) and that's not what the velocity of the car looks like....?

[@exaltedtoast6898]

I've been thinking of it as the sigmoid function (e^x)/(e^x+1) and I sometimes multiply that by C where C is just a random constant to make it more visible (I generally use 10)

[@traso56]

looks like YourMJKTube was right i.imgur.com/vYGIkQm.png

[@AnastasisGrammenos]

Has it been proven that you can NOT construct a function f(g(x)) witch takes a function g(x) as an input and has g'(x) as an output? And this is done "automatically". What I mean by automatically is that when you have a function lets say f(x) = x^2 - 2x plugging the value x = 3 gives you automatically the answer 9 - 6 = 3. So when i plug g(x) = (e^x) / (log(sin(sqrt(x^2/e^x)))) it will "automatically" give me the derivative as an answer. or can you?

[@jeffsamuelson7221]

Anast Gramm If I am understanding correctly, what you are describing is a more general kind of "function" whose input consists of the sort of functions with which we are more familiar; if that is the case, then the answer is "Absolutely!" In fancy mathematical parlance, the derivative can be regarded as a linear operator on a suitably chosen function space, such as the space of continuously differentiable functions. This operator (read: function of functions) would take something like f(x)=x^2 and return that function's derivative, in this case 2x. Notationally speaking, if we denote the operator by 'T', we may write T(f)=f'. Notice that T takes as input the entire function and not just the values at particular points. Function spaces: en.m.wikipedia.org/wiki/Function_space Differential operator: en.m.wikipedia.org/wiki/Differential_operator

[@franzluggin398]

More practically, every CAS (computer algebra system) that's worth anything can take an expression and a specified variable and compute the derivative. That output would then be g'(x) if the input was considered g(x). It's still in terms of x, but every CAS that's worth anything can have a substitution rule like x=3 applied to an expression.

[@tisajokt7676]

f(g,x) = lim(h -> 0+) (g(x+h) - g(x)) / h Or f(g) = { function(x) = lim(h -> 0+) (g(x+h) - g(x)) / h } It's just a matter of notation. Jeff's answer is better tho. :P

[@eduardogomes4865]

By automatically you mean closed-form. And no, you cannot have anything "more closed" than the actual limit definition of derivatives.

[@AnastasisGrammenos]

Thanks for the input! All of you!

[@Cosine_Wave]

An extra video... nice

[@leanderstephendsouza]

really loved it especially the jerk part, we're really taught this stuff in school

[@MelonMediaMedia]

For the weird people who want to know the ones after its in this order 1)Position 2)Displacement 3)Velocity 4)Acceleration 5)Jerk 6)Snap 7)Crackle 8)Pop

[@6funnys]

Not quite... while position and displacement are very much not the same, the shape of the graph is the same but with a possible upward or downward shift, being the initial position. Displacement is change in position, but not in reference to a change in time. Also, you would be better to write position/displacement as 0), as we tend to consider that as our basic function, our f(x). That way, you could label velocity, f'(x), its first derivative, as 1), then acceleration as 2) and so on.

[@ankeunruh7364]

Ask Tool to make an album!

[@krisbrandenberger544]

Hey, Professor Bertrand! So in general, for any Taylor polynomial, the coefficient c_n (the coefficient of x^n) controls the nth derivative of that polynomial evaluated at 0.

[@jameeztherandomguy5418]

???

[@prithvishah2618]

I love this channel so much Thank you so much

[@aldreivohna.aquino8191]

Very smooth and concise explanation!

[@edmilsonpoliveira2646]

I will translate the caption of this video into Portuguese. The video lessons from this channel are very good!!!

[@BenHeacox]

@ 04:27 The subtitles seem to not align with the audio at this point

[@TANUJKUMARPandey9999]

So i was studying the potential energy vs position graphy and there i encountered that second derivative of potential energy will give you the points of stable,unstable and neutral equilibrium. but now one told me how? So i searched the internet and youtube and here the search is end with this video.now i know why.so a heartfull thanks to creator of this video.your helping hand is changing the world in positive way.keep spreading love and knowledge.😊

[@yaskamuni9900]

1. Press Ctrl + Shift + I 2. Go to Console tab. 3. Copy Paste and press enter - const derivative = f => nth => x => { if(nth==1) return (f(x+0.0001)-f(x))/0.0001 ; else return (derivative(f)(nth-1)(x+0.0001)(0.0001) - derivative(f)(nth-1)(x)(0.0001))/0.0001 ;} 4. type and press enter - derivative(x=> x*x + x)(2)(1) 5. Gives you 2nd derivative of x^2 + x at x = 1.

[@aarjith2580]

*WAIT, Helpp! At **3:18** how did he get that d(df) is proportional to dx squared?*

[@well-being4443]

After watching your videos I felt if your channel were exist back in 2004 when I was a college students.

[@jordiegea7486]

Your videos are so cool. Love them 👌🏻

[@artur-rdc]

Rip me I watched the footnote after chapter 10 lol

[@sorinpanciuc5712]

same lol

[@danielparrado3605]

same asf lol

[@Re-nq2uh]

Brilliant video ✨ Thank you so much for it

[@jackkennedy98]

how can the gradient change more around the vertex of a parabola if it's second derivative is constant?

[@keithplayzstuff2424]

It was the same large amount all around.

[@franzluggin398]

He never compares the second derivative at 4 with the second derivative at other points of the same function. He just compares it to the shallower parabola.

[@nujuat]

jackkennedy98 it's large around the vertex... and everywhere else :P Unless it isn't a parabola, and then that just shows that even we use Taylor expansions in our head to approximate curves :P

[@avijeetjha8774]

3bb541 Keith since the 2nd derivative of parabola is constant so it shouldn't change while we move away from its curve but that's not possible since the rate of change of slope is decreasing.plz explain ☺

[@franzluggin398]

I have trouble understanding the question. Are you asking "How is it that the parabola has a very strong bend around its focus (its minimum/maximum) and less farther away from it?". Then the answer is: Because that's two different things. The slope of the tangent of x^2 is 2x. That means the slope increases linearly. What you see around the focus point is a very strong change in the _angle_ the tangent has with the x-axis. But the angle is the arctan of the slope, and so the farther out you go, the shallower the change in slope seems when you judge it by how much the angle changes.

[@Hercules003]

Everytime I see your videos I get a lightbulb moment. Suffice to say soon I wil run out of light bulbs to imagine lol. Thanks for the amazing videos.

[@avijeetjha8774]

Would anyone plz tell what derivatives greater than degree 2 mean mathematically like 2nd derivatives tells rate of change of slope then what does 3rd or 4th or nth derivative mean.

[@RetroGamingClashOfClans]

to push it a little farther 4th derivative of position vs. time is jounce

[@MikiSiguriči1389]

good man 3blue1brown

[@TheScienceNerdcom]

Wait a minute I am confused.....@1:50, the yellow linear function is referring to df/dx. It has a positive slope. So then he says the second derivative of the function would be 0 at x equals 4, or anywhere for that matter. This is definately wrong(RIGHT???). The second derivative would just be a nice horizontal line like y = 2 or y = 3 , whatever the slope of dy/dx is. Bc the second derivative is the slope of the firsts derivative. The first derivative is a line with a slope of lets say 3(that's about how steep his curve looks, whatever). So the second derivative is undoubatbly y = 3. I think I see the error. If the yellow line he drew actually was meant to represent f(x) (which should have been blue according to Grant's color scheme) , ONLY THEN will the second derivative be zero. Because if we start with a linear function as f(x), the first derivative will be a flat line, and then since the flat line has a derivative of zero, only now will the second derivative be zero. The mistake seems a little too obvious for me. Is there something I am missing?

[@SmithCS]

We worked with second derivatives all semester but I saw this notation on my calculus final and had no idea what it was.

[@balaportejean7015]

i love the small pi. Thx bro

[@parrychoi6350]

Can't wait for the next chapter

[@Re-nq2uh]

I've got a question I was wondering why is it dx^2? Why did we multiply the two dx? And i think it's because we need to divide the d(df) by the change in x that caused it namely dx the first time "and" dx the second, and since we're saying "and" (and "and" refers to multiplication) then we should multiply the two dx. I really want to know if what I'm saying is correct or it's something else that caused this.

[@isavenewspapers8890]

One way to think about it is to just think of a second derivative as, well, two derivatives. When you take the first derivative, that's df/dx, meaning that you've divided by dx once. Upon taking the second derivative, that's d(df/dx)/dx, so you're dividing by dx a second time. If that's what you mean, then you're right.

[@FourthDerivative]

This is my favorite video in the series so far, but I can't put my finger on why.

[@frictyfranq321]

Somebody explain me here please. Why is d(df)= (Some constant)(dx)^2 ? I mean the change in slopes is df2-df1 right? I don't understand how it is some constant - DX^2.

[@danielmarchionatti5746]

original: position velocity (1st) acceleration (2nd) jerk (3rd) snap / jounce (4th) crackle (5th) pop (6th) Lock (7th) Drop (8th) Shot (9th) Put (10th)

[@Krishna-xn8ss]

Thanks man this is so helpful

[@GourangaPL]

5:17 what's the music?

[@TheStarDreamer]

_If Displacement-Time graph of a ball moving, follows the function e^x exactly_ _Then, that is the most interesting type of motion in this Universe_

[@pratik2583]

Awesome work...!!!

[@1DInciner]

What function have constant second derivarive? circle?

[@robinswamidasan]

f(x) = x^2 ==> df/dx = 2x ==> d^2f/dx^2 = 2, which is constant. This is not true for a circle: x^2 + y^2 = r^2, using implicit differentiation (Chapter 6), we get: 2x.dx + 2y.dy = 0 ==> 2x.dx^2 + 2.dx + 2y.dy^2 + 2.dy = 0 (by Product Rule) ==> dy^2/dx^2 is not constant (because there are factors with x and y, and terms with dx and dy). You will get a similar answer if you use y = f(x) = (r^2 - x^2)^(1/2).

[@carultch]

No, a parabola has a constant second derivative. When we say the second derivative describes the curvature of a function, that is a first order approximation that is only true at low enough slopes where the first derivative's impact on the curvature is negligible. Both the first and second derivative need to be processed, in order to calculate the true curvature of a graph. Curvature as in the reciporcal of the radius of an osculating circle, that matches the function locally in position, slope and curvature.

[@irlshrek]

two videos in one day?! is it christmas already?!

[@18x9]

If the change in x or dx is doubled wouldn’t d(df) change by a factor of 2dx not (dx)^2. Why?

[@liviugheorghisan1130]

If the 2nd order derivative is positive, the function's graph "holds watter". If it's negative, it doesn't!

[@GeetanjaliVerma553]

Why we equate zero of minimum degree term in given equation to get tangent at vertex and why equate zero the coefficient of higer degree of x to find asymptote parallel to x axis Sorry sir these questions are out of this vedio bt if may possible so please solve my query.

[@ghostspyro9382]

And what about the integral of position in respect to time?

[@TheOiseau]

It does exist. It results in something called _absement_ which measures how far away something is from its initial position & for how long.

[@jameserayburn]

Another excellent video.

[@ominousscreech4054]

So intuitive !

[@Supware]

Will you be doing any videos on non-integer-th derivatives? Or is that too far removed from fundamental calculus..?

[@adityashankar5267]

Would it make sense to say that the 2nd derivative of 4 in a function is infinity

[@pikcube]

Not really. The 2nd derivative measures the speed at which the first derivative changes (which visually corresponds with how quickly the graph is curving). If we think of this in terms of limits, we can visualize what happens as a graph curvature speeds up. We eventually reach a point, as the curvature speed increases and we let the curvature approach infinity, where the graph makes an instantaneous turn, too fast to see the turning process. That is the closest a derivative comes to approaching infinity, although graphs in which the slope (and therefore the curvature) changes on a dime are said to be "not differentiable". This means that our approach of taking tiny nudges to see how slope or curvature changes fails to capture any meaningful information. A concrete tiny nudge, no matter how tiny, doesn't give a meaningful approximation of the slope or curvature. Because of this, the derivative or second derivative at these points are said to simply "not exist". That being said, that doesn't make the idea of an infinite derivative wrong or bad, math is all about taking these fuzzy ideas and playing with them until they make sense. Mess around with it, try to set up what infinite slope or infinite curvature looks like, and see what it teaches you.

[@Александр-л8з3э]

Amazing explanation !!!

[@ioangauss]

Great for students animations rocks !!!

[@mukhtaarjaamac8763]

Integration by substitution Non added but it is chain rule integrated

[@anonymoose3423]

If higher-ordered derivatives are meant to represent the curvature of the graph, how is the second-order derivative of a function at its inflection point zero ? Thanks in advance.

[@cookieshade197]

Because, at the inflection point, the curve is flat (or approches flatness, atleast), and so, the curvature ought to be zero. Make sense?

[@anonymoose3423]

CookieShade Yes. Thanks!

[@gvarph7212]

I've first learned derivatives years ago, but I've only just figured out how the (df/dx notation works). For some reason, I've always thought that d^2 f / dx^2 was d^2 f / d (x^2) and that just made no sense to me

[@pratisthatiwari9101]

Thank you😊

[@凌嘉徽]

❤Helps a lot,love from China🎉

[@devrajyaguru2271]

thank you for this great video

[@bayubetaB]

ah if only you had posted this video when i was taking calculus class in my freshman year

[@isobar5857]

3blue1Brown. This always confuses me in the following sense. Why does it necessarily follow that if the second derivative is decreasing it has to be negative ?

[@isavenewspapers8890]

That's not true. Maybe you meant that a decreasing first derivative means a negative second derivative, or a decreasing second derivative means a negative third derivative. The reason for this is that each subsequent derivative tells you the rate of change of the previous one. That is precisely what the derivative is meant to measure. For example, if you have negative acceleration, such as from slamming on the brakes of a car, that means that your velocity is decreasing.

[@elen1ap]

Where did you get the function of the distance of the car in terms of time?

[@demian893]

why the change of the change, the second derivate, is proportional to dx^2 since dx^2 could be discarded because it's so small, like in the other videos?

[@nikhilnagaria2672]

It can be discarded in comparison to df, but here you have d(df), so that's again very small, and comparable to (dx)^2.

[@nikhilkamble4210]

Great video👍. Can you make videos on optimization with linear programming?

[@joshuaronisjr]

Why is the difference of the differences proportional to dx^2 ?

[@fountainovaphilosopher8112]

Is it possible to have a derivative of a non-integer order or something? And if it is, how can it be explained?

[@zairaner1489]

en.wikipedia.org/wiki/Fractional_calculus

[@fountainovaphilosopher8112]

Thanks bro.