Continuity vs Partial Derivatives vs Differentiability | My Favorite Multivariable Function

Рет қаралды 63,093

Күн бұрын

In single variable calculus, a differentiable function is necessarily continuous (and thus conversely a discontinuous function is not differentiable). In multivariable calculus, you might expect a similar relationship with partial derivatives and continuity, but it turns out this is not the case! In this example, what I call the cross function, we will see that both partial derivatives exist but that the function is nonetheless not continuous. This means our work is still cut out for us: we need to define a new concept to replace single variable differentiation in the multivariable context that is more than just partial derivatives existing.
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Пікірлер: 75

@aquilazyy1125 3 жыл бұрын

I really like how you connect concepts to each other like this, which is the hardest part of learning anything, but is what most other teacher fail to do. Keep up the great work!

@DrTrefor 3 жыл бұрын

Thank you! I agree that is the hard part but also the most important!

@aquilazyy1125 3 жыл бұрын

Dr. Trefor Bazett I checked out your other videos and they are really inspiring. This will be a fine addition to my collection :P

@travorreznik7295 3 жыл бұрын

The idea of taking multiple concepts and showing their connections to each other is awesome! Very useful and informative video. Thanks a lot.

@DrTrefor 3 жыл бұрын

Glad it helped!

@mathCS1233 2 жыл бұрын

you just helped me solve a sum that I was trying to do for some time now. thanks a lot!!

@at6727 3 жыл бұрын

video gr8 editing gr8 explanation gr8.. you are gr8!

@briandwi2504 Жыл бұрын

Lovely and helpful example.

@continnum_radhe-radhe 2 жыл бұрын

Thank you so much sir 🔥🔥🔥

@not_intrested 3 жыл бұрын

God bless you man ♥️

@ahmedtawfeeq6511 4 жыл бұрын

Amazing teaching method! Keep it up, please.

@beoptimistic5853 3 жыл бұрын

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@maph420 5 ай бұрын

really useful playlist :)

@apoorvajain9137 2 жыл бұрын

sir can we take partial derivative with respect to x by keeping y constant even if the function is non differentiable? or do we need to follow the defination?

@citiesinspace4864 4 жыл бұрын

Hey there professor, just curious but what software do you use to draw your graphics? Is that a result of a graphics library from a programming language?

@beoptimistic5853 3 жыл бұрын

kzfaq.info/get/bejne/jrd5ldFjnMqzmnk.html 💐💐👍

@GovindKumar-mu6jq 2 жыл бұрын

Thanks ❤

@srinidhir2188 3 жыл бұрын

ULTIMATE EXPLAINATION SIR 🎉 FROM INDIA 😊

@DrTrefor 3 жыл бұрын

Glad you enjoyed!

@shrutigupta584 Жыл бұрын

What would be the partial derivative wrt y at point (2,0) .. will it exist?

@amansingh-ww2qc 3 жыл бұрын

Love u sir

@abrahamsweetvoice7687 4 жыл бұрын

My calc 3 exam is tomorrow, these vids are saving me !! ( also the divergence theorem is killing me rn xD ) Are you thinking of making videos about the greens theorem, divergence theorem, line integrals,... ?

@duckymomo7935 4 жыл бұрын

Of course: those are the fundamental theorems of Vector Calculus, not covering them would be a crime Also check out Michael Penn

@spyninja1456 3 жыл бұрын

Thank you for this video. What happens regarding the partial derivative for the rest of the function (where x and y do not both equal 0)?

@rizalpurnawan3796 3 жыл бұрын

I guess it equals zero as well. As that step function forms planes in x-y coordinates at xy does not equal zero. Hence if we take y equals to some constant c, we see that it's some kind of constant line along x at y=c. The derivative of a constant line is equal to zero, hence the partial derivative of f(x,y) with respect to x is equal to zero, as well as the partial derivative of f(x,y) with respect to y is equal to zero. This is the characteristic of step function.

@jamesjin1668 3 жыл бұрын

I agree that it's zero. If you choose any point on the rest of the function, you will soon discover that no matter what h is, the output of the function simply doesn't change. Therefore, as you derive by first principle (using limit), the numerator will simply be zero, similar to the partial derivatives at (0,0).

@beoptimistic5853 3 жыл бұрын

kzfaq.info/get/bejne/jrd5ldFjnMqzmnk.html 💐💐👍

@toniivanov4419 2 жыл бұрын

Mistake at 5:15. It should be "unfortunately." Thanks.

@sanjaydwivedi7167 3 жыл бұрын

Sir, how to prove that partial derivatives of a function exist..

@CstriderNNS 4 жыл бұрын

is the reason we evaluate a limit up to but not including a function due to the fact that at the funcion the secant line goes to infinity for all functions ?

@beoptimistic5853 3 жыл бұрын

kzfaq.info/get/bejne/jrd5ldFjnMqzmnk.html 💐💐👍

@lanog40 3 жыл бұрын

Dr. Trefor Bazett, have you joined the 3b1b "Summer of Math Exposition 1" discord? It could use more math youtubers, especially of your caliber.

@paultayar6281 3 жыл бұрын

Hi. Does the directional derivative in v(1,1) or any other vector different than x or y axis exist at (0,0) for the function f(x, y) =0 for xy different than 0 and f(x, y) =1 for xy = 0 F(0+s.1,0+s.1) - f(0,0) =0-1 while 1.partialx +1.partialy =0

@DrTrefor 3 жыл бұрын

Nope, none of the other direction derivatives exist because you move one step in any other direction and you go over a step in the function

@1aadithya9c25 3 жыл бұрын

Superb animation sir

@DrTrefor 3 жыл бұрын

Thank you!

@saimsiddiqui1544 3 жыл бұрын

How cool you are😍😍😍

@Jack_Callcott_AU 2 жыл бұрын

A question: A couple of videos back, in this series ,you asked the question :" If , in the case of a function z=f(x,y) we want to calculate the limit (x,y)--> (a,b) , is it sufficient for the existence of a limit L to approach (a,b) along every possible straight line and get the value L for the limit to exist." I say it is. Am I right ? It is clearly a necessary condition, but is it sufficient.

@DrTrefor 2 жыл бұрын

It is not sufficient! We have examples where along every straight line the limit is L but along some weird curves that are not lines the limit is something else.

@Jack_Callcott_AU 2 жыл бұрын

@@DrTrefor Thanks for the help. I'm working my way through your multi-variable calculus course. I'm up to video 18 and I'm really enjoying it.

@vijaysinghchauhan7079 2 жыл бұрын

When we define the partial derivative of f(x, y) wrt y we usually use k instead of h, but why?

@carultch Жыл бұрын

Possibly because h is already spoken-for, for the same action when defining the partial derivative with respect to x. We therefore choose k. Why k of all possible letters? It probably has to do with the convention of using the point (h, k) to identify the vertex of a parabola, or the center of other conic sections. The h is used for the x-direction horizontal shift, and the k is used as the y-direction shift. I don't know why we use k specifically for that, but it probably is due to the fact that the letters i and j are avoided, since these letters likely might refer to the imaginary unit. It's common that c and k are used interchangeably to stand for constant, since konstant is the German word for constant.

@gamersfx7167 2 жыл бұрын

If a fn is continuous than it is necessary that partial derivatives exist and will have same value sir?? Is that so plz reply sir If no can u plz give me a counter example for that plx sir🙏

@DrTrefor 2 жыл бұрын

Sadly that isn't true:( Just take any function with a point somewhere.

@gamersfx7167 2 жыл бұрын

@@DrTrefor sir what about the diffrentiation and parial derivative 1.)Can I say if a fn is diffrentiable than partial derivative exist at that point and must be equal.. Or it is also wrong is it 2.)possible to get a fn which is diffrentiable but parial derivative does not exist at that point?? Is that so?? Which one is crrct 1st and 2nd plz reply sir and thanks for clearing my first doubt sir ❤

@atirmahmood7058 9 ай бұрын

Actually for partial derivatives to exist the function should be continuous along the axis even if it is not continuous throughout the disc centred at (a,b).

@claudiolopez9794 2 жыл бұрын

lov u

@premsubhrajeetbhoi7259 2 жыл бұрын

is differentibility and differentiation are same? sir

@carultch Жыл бұрын

Differentiation is the name of a process. The process of taking a derivative. Derivative is a noun, differentiate is a verb, differentiation is a noun, referring to the process of that verb. Differentiability is a noun for the concept of the adjective differentiable, which describes a function for whether its derivative is continuous or not. If the derivative of a function is continuous, it means it is differentiable. If there are jump discontinuities in the derivative, or places where the derivative has a vertical asymptote, then it isn't differentiable at that point. You will see a kink or a cusp in the original function where this happens, which is a point where it is not possible to define the derivative in introductory Calculus.

@achalcharantimath5603 4 жыл бұрын

please make a video on gradient decent

@beoptimistic5853 3 жыл бұрын

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@chetanbhatt5153 3 жыл бұрын

🤩

@KevHnh 4 жыл бұрын

Mr. Bazett, please end my curiosity. Are you actually on Bad Vibes Forever?

@KevHnh 4 жыл бұрын

Trefor Bazett I find your discrete math videos easier to understand than that response 😭

@duckymomo7935 4 жыл бұрын

with multivariable it does get complicated Thm1 Separately continuous not implies continuous Thm2 Continuous in every linear direction does not imply continuous Thm3 Existence of partial derivatives does not imply differentiable Thm4 Existence of directional derivatives in every direction not implies differentiable

@beoptimistic5853 3 жыл бұрын

kzfaq.info/get/bejne/jrd5ldFjnMqzmnk.html 💐💐👍

@ManishSingh-gc5fv Жыл бұрын

What about the Dirac delta which is the derivative of Heaviside function?

@carultch Жыл бұрын

That's a special exception, where a derivative is defined despite a jump discontinuity. The Heaviside step function is technically not differentiable at x=0. We just define the Dirac impulse function to be its derivative, in order to keep track of the calculus implications of discontinuous functions. It works well with the Laplace and Fourier transforms, but it is difficult to make these functions work for introductory Calculus methods.

@Happy_Abe 3 жыл бұрын

I think something possibly more powerful we can say than this but I'm not positive is: Partial derivatives can exist, be continuous, and themselves differentiable; all while the original function is not continuous. Is that correct?

@DrTrefor 3 жыл бұрын

Actually no. Continuous partial derivatives implies the function is differentiable. This is a theorem, but is non-trivial.

@Happy_Abe 3 жыл бұрын

@@DrTrefor So I’m aware of this theorem but it has an extra condition That is that the function is continuous as well Theorem states precisely that if (a,b) is in your domain of f and f(x,y), f_x(x,y), and f_y(x,y) are all defined in a neighborhood around (a,b) and are continuous in that neighborhood then f(x,y) is differentiable at (a,b).

@carultch Жыл бұрын

@@Happy_Abe If a function is not continuous, it is not differentiable either. A jump discontinuity in the original function, will mean that the derivative will spike to either infinity or negative infinity at that point. This means that there is a jump discontinuity in the derivative as well. You might say that the Dirac Delta function for a unit impulse is enough to make the Heaviside step function differentiable, but it really isn't. That is more of a work-around to get the Calculus to work, despite a singularity. These singularity functions work well for Laplace and Fourier transforms, but are very difficult to make work with introductory Calculus methods.

@Happy_Abe Жыл бұрын

@@carultch but does a function being discontinuous imply it doesn’t have partial derivatives defined for its whole domain? Otherwise the extra condition seems unnecessary

@carultch Жыл бұрын

@@Happy_Abe Yes. Discontinuities imply non-differentiable functions.

@aizazhashmi9433 4 жыл бұрын

Please make an other video on limits of multivariable function using epsilon delta definition.. How to prove that a limit exist or doesn't exist using epsilon delta definition.

@duckymomo7935 4 жыл бұрын

It’s easy to prove that it doesn’t exist Proving existence is harder

@beoptimistic5853 3 жыл бұрын

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@rizalpurnawan3796 3 жыл бұрын

@@duckymomo7935 we can consider continuous function on metric space as such a function is the generalization of continuous real-valued multivariable function, i.e., we consider the metric space (R^n, d*) with d* is the standard Euclidean distance. The function can be given by f : U --> R, where U is the subset of R^n, hence we consider f as a function between metric spaces (R^n, d*) and (R, d), with d is defined by d(x, y) = |x - y|, for all x,y in R. If f is continuous at a point u in U, then for all ε > 0 there exists some δ > 0 for all x in U [ d*(x, u) < δ => d(f(x), f(u)) = |f(x) - f(u)| < ε ] must be satisfied. Q. E. D.

@rizalpurnawan3796 3 жыл бұрын

But, it is out of context. This definition doesn't occur in elementary multivariable calculus class. We might find it in General Topology class.