What is differentiability for multivariable functions??

[@DrTrefor]

This video actually defines differentiability too strongly. I introduced two error functions. However, I incorrectly said those error functions were only functions of x and y respectively. Both should be functions of BOTH variables and small when the limit as BOTH variables goes to zero.

[@manuelkarner8746]

so when x goes to 0 & than y goes to 0 the function is not differentiable but when both variables go parallel (?I am confused) to 0 than the function is differeniable ? so there is a kind of a "harder to detect" differentiability here ? (I messed something up, didn´t I ?)

[@AangContreras]

So... In the video they are defined as: E_1 = E_1(Δx), E_2 = E_2 (Δy). Do you mean that in this particular case, they should be: E_1 = E_1(Δx, 0), E_2 = E_2 (Δy, 0) ? Because we where computing the partial derivatives just in (0,0). Could you explain those error functions with a little more detail? Maybe in another video of continuity of a function and not just at a point...? Still if you don't have the time to make it, I really thank you for your videos! You are one of the best teachers on KZbin

[@jamesjin1668]

@@manuelkarner8746 Just talking about the general definition of differentiability, the overall notion is that the true difference between two outputs f(x0,y1) and f(x1,y1) should be very closely estimated by the difference between f(x0, y0) and f(x1, y0) plus the difference between f(x0, y0) and f(x0, y1). The error terms are there to make up for the inaccuracy, but in order for the function to be differentiable, as x1 - x0 and y1 - y0 approaches zero, the errors must still be very small compared to the change in x and y. I wasn't sure what you don't understand exactly, I hope this explanation helps.

[@punditgi]

Can you please reshoot this video. It would be nice to fix the error and also to emphasize the linearity of the derivative on both cases (single variable and multiple variables) and also prove that the limit of the difference quotient and the expression with the limit of the error term are equivalent .

[@danielchin1259]

This is a very important correction.

[@dAntony1]

I don't understand why this 15 minute video is more clear and concise than my 1 hour 50 minute lecture that occurs 2xs per week.

[@DrTrefor]

That must be so frustrating to attend that lecture!

[@amilagoku6073]

also thinking that for two years now. And also gonna think that for two years from now.

[@crewberggren7669]

If the Uni lecture is bad nothing really happens to the professor. If this video is bad, we immediately leave and look for another video

[@toomanycharacter]

Well, KZbin creates a competitive environment for videos, so when you search "what is differentiability in multivariable calculus" you are likely to find the most popular video. For science-related topics, this video is usually also the easiest-to-understand. You can easily just watch a different channel on KZbin, and KZbin has just so much content. So you can find the easiest-to-understand, best-quality content for free on youtube. Universities don't get to have such a unique competitive scene. You (usually) can't just choose to listen to a different professor's lectures without going to a different university. Which is why you don't (usually) get the highest-quality content from a university.

[@blugreen99]

A good lectue== no chalk no distracting handwriting movement. Body and hands never block view of text and diagrams which are well coloured with shading and 3d effects . Color used on text also. Do we need to see lecturer all the time? 3 blue one brown uses similar to Trevor but is not visible all the time?? The perfect lecture technoque??

[@ritxve]

Wow! best explanation of differentiability in multivariable calculus, I have never seen someone explaining single variable derivative in that E(h) form, this helped me to relate quickly E1 & E2 terms. Thank u very much professor🙏

[@winmintun2992]

Yes the best

[@praveshkumar8267]

This is one of the easiest explanation i have seen for this. Thank you

[@DrTrefor]

Glad it was helpful!

[@Anonymous-vo8ks]

@@DrTrefor thanks dude.

[@ozgeylmaz8685]

This video was amazing, it brought tears to my eyes! Now, I can view the world from various perspectives thanks to what I've learned.

[@punditgi]

It would be nice to do the epsilon-delta proof that limit exists for the difference quotient for the single variable case, then do the same for the alternate formula with an error term, then do the same for the multi variable function. Finally, prove the differentiable theorem for multi variable functions with continuous partial derivatives.

[@abmohit3117]

outstanding and amazing ... the reason why i love calculus ...and the reason many don't love calculus becuz we don't have the teachers that can show the real beauty of calculus in this way...Really appreciated and very happy that i got someone who teaches the way i want to learn the beautiful things ...such as calculus !!! lots of love sr ...✨✨

[@giacomogavelli4095]

Thank you for this video, it gave me an explication to a problem I found doing complex analisys and that's really cool

[@chauhan.narendra]

Thanks a lot for explaining the concept so clearly. I have two questions for you : (1) Why did you choose to add both of the equations containing the E₁ and E₂ when we move from (x₀, y₀) to (x₀+∆x, y₀+∆y) at 08:52 ? (2) Can you please explain the theorem stated at 13:30 ?

[@tanjamikovic2739]

I have same questions. (1) How so that E1 and E2 is the same in both equations. (2) where did the continuity come from?

[@HelloWorld-dq5pn]

Have the same question.

[@samyakgupta4301]

Sincerely thank you.

[@_mippi_]

At 9:23-9:26: small typo for the error term for y. You have E_1 and E_2 as different functions. Not sure if this mattered but I understand what is going on. Fantastic video! :)

[@amansingh-ww2qc]

No one can teach like u sir ,love u sir

[@sunraiii]

You're a fricking genius, m8

[@darthglowball]

@4:38 I understand that E(h) is a relative error, but what is E(h)/h? A secant line of the error function with its left vertex at (0,0), where its right vertex approaches (0,0)? Why would you need that in the limit, while we already know that the error approaches zero when h --> 0?

[@darthglowball]

Wait, maybe the answer is that if you have a limit of E(h) it detects differentiability most of the time, but having a limit of E(h)/h catches corner cases, literally! Like for an absolute function f, where both lines can't occupy the same corner, you have to choose which line occupies it, and when shooting a tangent from that corner along the chosen line, if you use E(h), you would not be able to detect differentiability because the limit of the error (tangent line minus a line of f) is zero while we have a corner! So to detect the corner, we look at the derivative of the error, because that is a constant, which dissatisfies the limit equation! Basically, you look at how fast the moving points that define the error hit each other at (0,0)! A soft hit of the points (derivative of error = 0) means there's no corner! A hit with a high velocity means there's a corner! Maybe there's even more cases that it catches?

[@mrslave41]

this confused me too. the solution to the confusion is to divide both sides by h. which is just the original equation. but without the limit. instead, you now add the error term E(h)/h. and now you say that it goes to zero as h->0. lim h->0 [E(h)/h]=0. saying that lim h->0 [E(h)]=0 is not enough. Because lim h->0 [E(h)/h] might not equal zero.

[@douglas5260]

Thinking about the derivative when walking along a diagonal path between the X axis and the Y axis. Example (using a large step to be clear, but in reality thinking about a much smaller one), from (0, 0) -> (1, 1). In the video, you decomposed the resulting derivative as if we were walking a step in the X direction, and then a step on the Y direction. You then added the partials and the respective errors. If we first move in the X direction from (0, 0) -> (1, 0), we add the partial X at (0, 0) * delta X, with the error caused by delta X. And when moving from (1, 0) -> (1, 1), we add the partial Y at (0, 0) * delta Y, with the error caused by delta Y. But in the second step, since we are now moving from (1, 0) and not from (0, 0), shouldn't the Y derivative not be evaluated at (0, 0), but at (1, 0)? In other words, dF/dy on the equation at 10:38 should be evaluated at the point (X0 + delta X, Y0), and not (X0, Y0)?

[@amansingh-ww2qc]

Sir u teach so well that I love u

[@julioreyram]

Nice intuitions and visualizations. Thanks!

[@Bermatematika]

Usually to define differentiability for function of two variables, I try to convey it geometrically. I say that analogous to the one dimensional case, to be differentiable meant that we want to have a plane that approximate the surface "well enough". How we define this notion of well enough? If we move (x,y) to (x+delta x, y+delta y) the error of the z value between the surface and the plane the relative error of the z values between the surface and the plane with respect to the displacement from (x,y) to (x,+ delta x, y+ delta y) is goes to zero.

[@Bermatematika]

@@DrTrefor Great! Just to let you know, for me I feel your audio is not as crisp as your video. But it seems no body comment about that. If no one else agree with this, just ignore my comment. Peace!

[@gurulinggbiradar6982]

6:17 love that graph

[@lolololxd456]

sos un CAPO increible maestro muchas gracias

[@Reptilian.cricket]

really great video and whole playlist!!! Thank you :)

[@viveksingh8823]

Thank you so much for the gift🙏🙏

[@KidsCastable]

I dont understand why you can just add the 2 partial derivatives in the way you did at 8:49. You are saying that: [ f(x + dx, y) - f(x, y) ] + [ f(x, y + dy) - f(x,y) ] = f(x + dx, y + dy) - f(x, y) but that would mean f(x + dx, y) + f(x, y + dy) - 2f(x, y) = f(x + dx, y + dy) - f(x, y) I don't see how this is obvious or even allowed. Could someone help me understand why we can just add the 2 partial derivatives?

[@rizalpurnawan23]

Oh I see, so the function f is differentiable if the total derivative exists, am I correct? The expression in min 9:09: delf/delx • ∆x + delf/dely • ∆y + E1(∆x) + E2(∆y) As we take limit (∆x, ∆y) --> (0,0), it becomes delf/delx • dx + delf/dely • dy which is precisely the total derivative Df(x,y), am I correct?

[@023aashish]

I guess he only means the secant to be almost equal to the tangent. i.e. ∆f almost equal to df. Total derivative does not use ∆f .

[@sayanjitb]

you can say that is the total differential of function f(x,y).

[@benjaminyellin5095]

Hey Dr. Trefor, absolutely love your videos! Quick question: Per the last theorem, it would seem that the cross function is differentiable - when we proved it isn't. What am I missing? Thanks!

[@DrTrefor]

Correct, it is not differentiable. It is not even continuous which is a necessary condition for differentiable. What is confusing is the PARTIAL derivatives do exist, but that isn't sufficient for differentiability.

[@vicentesepulvedatrivelli5843]

@@DrTrefor but aren't also the partial derivatives continuous, aso both are 0?

[@jacobfertleman1980]

@@vicentesepulvedatrivelli5843 no - as it is an open region - therefore the open region with zero must contain a point (t,t) and in the direction of x it jumps from zero to one

[@matanshtepel1230]

great video! will share with my classmates 😁👍

[@AkhileshYadav-ol7jy]

A tremendous explanation Thank you sir Love ❤️ form india

[@muhammedyusufsener1622]

Excellent Sir, thanks!

[@adityaghosh9955]

Brilliant video!!!!!!!!!!!!!

[@SanjinHalilhodzic-ny2zj]

Great video, just one question. What is considered small for the error terms?

[@jakebrowning2373]

So does the last theorem shown only apply in specific cases? Since the theorem failed in the "cross" example

[@rioda5971]

I think the theorem applies to the cross example too. If you take an open region of R^2 that contains {(0,0)}, you always will have an element {(0,y)} in the open such as y≠0. Then, if you try to compute the partial derivate of f respect x in (0,y), you will find that doesn't exist, so the partial derivate function isn't continuous either.

[@cooking60210]

This is a kind of weird definition of differentiabiility but it's the standard one given in books that don't assume linear algebra. Using linear algebra there's a much more conceptually satisfying definition.

[@VectorSpace33]

@9:29 He has E_1 multiplying delta y. It should be E_2.

[@aslgulyalcndag318]

i didn't fully understand how to understand if a partial derivate is continuous. in this video we said that this func. has partial derivates but it is not differentiable. so how we can apply the last theorem? thanks in advance

[@yuxuantian1182]

I think it's because the example given has partial derivatives at (0,0) instead of in the region of (R,R). For example, if you look at the point (1,0), dz/dy does not exist since it's discontinuous.

[@ashwinraj5215]

great work

[@SumitKumar-ne9kc]

i do not understand this last theorum at 13.44: if we apply this theorum in our pervious example (in which we make a cross ) then we find that df/dx and df/dy is continous so acc. to this theorum function should be diffrentiable in region R . Sumit Kumar 1 second ago but we already prove that it is discontinuous .pleeeeeeeeeeease expain it

[@mateoruiz7239]

Great video!

[@amansuryavanshi8748]

Sir great video!

[@Mark-dc1su]

This kind of reminds me of the squeeze theorem in a way. The assumption being that if something has a partial derivative in the x and y directions, there should be a point between the x and y axis where the function is defined. Is this a valid way of understanding differentiability?

[@dwaipayansharma2282]

This is awesome

[@arnavbhavsar8567]

In this case if I go to find partial derivative of x and y at (1,1) or any point not on axes, it will be 0 for both the cases. Also the partial derivative on any of the axes is also 0 for both coordinates. Then this function should be differentiable. But that as you said, it is not differentiable. Help me with this

[@tulio829]

It's differentiable on point (1,1) and on other points except on the points located on the x and y axis. So, OVERALL, f is not differentiable because is not differentiable in every point of its domain.

[@Festus2022]

Why is that in a typical multivariable function f(x,y) you DON'T add the x and y partial derivatives to get a full derivative, but in a parameterized multivariable function ( eg. with time as the independent variable and x and y as the intermediate variables) you DO add the x and y partial derivatives times their respective full derivatives with respect to time ( dx/dt and dy/dt)?

[@AllRounder-jt1pe]

our prof also refers your vedios. Thanks a lot.

[@thebrainfeed402]

looks like you are from IIT Jodhpur😂😂😂

[@AllRounder-jt1pe]

@@thebrainfeed402 B20CI026 , How r u bro 😅🤟

[@thebrainfeed402]

@@AllRounder-jt1pe F roll no. bhi leke aa gaya😂😂😂

[@AllRounder-jt1pe]

@@thebrainfeed402 tm kaho to shi tmhari poori janam kundali khol ke rakh denge .😅

[@thebrainfeed402]

@@AllRounder-jt1pe chal keh diya

[@abdulghanialmasri5550]

Another great video, many thanks. Would you consider more topics in numerical analysis, like stability, interpolation, extrapolation, gaussian quadrature, etc.

[@sciencewithali4916]

Thank you so much professor!!

[@kaagaj_bottle]

i am happy that this video has 0 dislikes

[@khalilmohammed2297]

thank you so much and i have a question why don't you put the error as (delta y -dy) i have not understand why you made E(h) .

[@sarath3688]

Sir, the theorem in summary is necessary and sufficient condition for differentiability, right?

[@shalinishankar2762]

Sir, is the theorem only a necessary condition for a function to be differentiable?

[@garlicat878]

Very nice video! It helps a lot. I have a question about the error terms. I saw it's written in the form like sqrt((deltaX)^2+(deltaY)^2) somewhere else, and then it put the equation back to the limit equation form, check whether the limit exits to check if the function is differentiable. Does it make sense? How should we deal with the error terms when solving real problems?

[@DrTrefor]

For specific types of functions, you can actually compute the formula and try and show the limits are zero, but the algebra for that will depend on each specific function

[@gainknowledgewithgeetu8829]

How do u made your presentation? Which software u are using

[@makeitreal777]

A heartfull thank you sir

[@Rohan_Mahato]

Very nice

[@artistiaay5512]

So when u say h goes to zero ,,that never means h is equal to zero ! it just gets closer to zero without actually being zero?

[@DrTrefor]

Exactly. A limit is about things ARROUND a point, not actually AT the point.

[@abhishekpg9615]

@@DrTrefor That should also mean that we are theoretically taking the slope of a secant line (since h is not equal to zero) while taking derivative instead of tangent line.

[@GSurendra9119]

super duperrrrrrrrrrrrrrrrrrrrrrrrrrrrr sir thank you sir

[@SajjadAhmad-wv8mw]

Video was amazing, but how we can know that 'f_x' and 'f_y' are continous on that region.?

[@carultch]

Similar concept as continuity in general. The limit of the function as both x and y approach each point must equal the value of the function, in order for the function to be continuous. And you need to get a consistent answer no matter what path you take to the point, for the limit to exist. As an example of a function that isn't continuous, is f(x,y) = x^y at the point where x=0 and y=0. You get conflicting answers, depending on how you approach this point. Hold x constant, and the limit of this function is 0. Hold y constant, and the limit of this function is 1. Approach along the line y=x, and the limit of this function is 1. You get conflicting answers, which means the limit does not exist. You will see a jump discontinuity as a consequence of this, if you graph this function in 3-D.

[@GoogleUser-ee8ro]

interesting how partials exist but the function is not differentiable. That reminds me of some theorem (maybe the same) in complex analysis, of a function being analytic, it needs some sort of "neighborhood", another way to say continuous i guess.

[@munyanezajeannepomuscene9029]

We need clear examples that can help us to understand how use those formulas

[@serkanbasatlk3322]

\[ f(x,y)= \left \{ \begin{array}{ll} \frac{xy^2}{x^2+y^2} & \text{ if } (x,y) eq (0,0)\\ 0 & \text{ if } (x,y)= (0,0) \end{array} ight . \] to show that this function is not differentiable at (0,0), is it really enough to show that its first order partial derivative with respect to x or y is discontinuous at (0,0)

[@AbjSir]

Thanks sir

[@keithchan379]

When you use this condition to check if a function f is differentiable, you first have to be able to write down the partial derivative fx and fy??! But being able to write down the partial derivative means we have assumed f is differentiable in the first place? Is this not circular reasoning?

[@DrTrefor]

Not quite. Differentiability is a STRONGER condition than just the partial derivatives existing. So yes, if those don't exist, it definitely isn't differentiable. But if they do, you still need more work to how full differentiability

[@keithchan379]

@@DrTrefor Thank you for your reply! That clears my concept :)

[@mustafanaser9789]

I will finish hopefully my undergraduate in 2 years. I would then appreciate your work by donating some money. Is it possible?

[@DrTrefor]

Hi Mustafa, thank you so much and I would always love people becoming channel members, just click the join button under any video:)

[@nasirjoya1450]

i would love a video with a bit of slower speaking speed, cause i am lost with that speed of speaking, i catch one thing and then the second is gone, Respect

[@sayanjitb]

Dear sir, for this pathological example, as defined in the video, the partial differential ceases to exist along path y=x. Hence according to the last theorem, partial derivatives are not continuous; hence function is not differentiable. Am I right?

[@sayanjitb]

Please anyone help me out!

[@continnum_radhe-radhe]

🔥🔥🔥

[@engincangocer9962]

perfect

[@farukyldrm8498]

saygılar reis. anladım.

[@thomascanellas5476]

Thank you!!!!!

[@joejoeman3645]

luv u mane

[@justanotherguy469]

Your error term is already embedded in the concept of limits. I can not see its necessity.

[@vijaysinghchauhan7079]

The error function i.e. E(h) is not very intuitive to me. Anyone here has the capability to put the concept in a way that is nearly intuitive?

[@dhh1486]

studying mat235 and came across ur channel 😂😂

[@DrTrefor]

Any chance your at university of Toronto? I know their multicariable calc course is 235

[@dhh1486]

@@DrTrefor yeah were u a TA once upon a time for that course?

[@mariamyounis9768]

Who is unliking these videos, I’m so mad >.

[@DrTrefor]

me too!!!!

[@madhavvenkatesh6782]

🐐

[@gagansoni8185]

Here in that example both the derivatives are zero ie constant => continuous so according to the theorem function should be differentiable... Contradiction 🙄🙄🙄

[@shuriken188]

The partial derivative with respect to x along the x axis remains 0, yes, but along that line, the partial derivative with respect to y is only 0 at the origin. For every other point along the line, the partial derivative with respect to y does not exist as the function is not differentiable by the limit definition of the single variable derivative: As dy goes to 0 from the negative side: f(1,dy)-f(1,0) = 0 - 1 = -1 1/dy goes to negative infinity Therefore the limit from that side is positive infinity (as you get closer to the discontinuity, the slope between your point and the value at the discontinuity approaches infinity) As dy goes to 0 from the positive side: f(1,dy)-f(1,0) = 0 - 1 = -1 1/dy goes to infinity Therefore the limit from that side is negative infinity (as you get closer to the discontinuity, the slope between your point and the value at the discontinuity approaches negative infinity) Algebraically, the limit does not exist so neither does the derivative. Geometrically, as you get closer to the discontinuity, the slopes do not approach each other, so there is no one value that describes how the function changes at that point, thus no derivative. Therefore, the partial derivative of f(x,y) at (1,0) does not exist and the partial derivatives do not exist for all (x,y) Therefore the function is not necessarily differentiable at all points.

[@-baxtyarmaths4556]

🌷

[@kiarash7604]

Wow, there is some math between all these ads

[@JohnMoriarty-rr4yg]

❤

[@userzcxtgc7129]

Sir, What is your Religion?? Can I know?? Please reply as soon as possible.