Differentiability for Multivariable Functions

[@arnavgupta7392]

Simply one of the best videos on differentiability of 3 functions with 3 variables.

[@isharauditha4257]

the best explanation I have found on youtube about multivariable differentiation

[@glaveglave7000]

Your video is easy to understand, simple, clear, detailed, overall excellent

[@YonkoCraft]

Good simple and efficient i love it ♥️

[@tanvirfarhan5585]

good explaination

[@sezgiselek]

u the best love it!!!🤩

[@gauthierruberti8065]

Thank you so much!

[@ver-li-o]

I think the plane in the limit should be L(a,b) and not L(x,y), since you are measuring the distance between the image of (x,y), i.e. f(x,y), to the plane tangent to the point (a,b), i.e. L(a,b), as (x,y) approach (a,b).

[@mathemation]

Thanks for your comment, Oliver. It's a good thought, but not quite correct. The quotient in the limit definition needs L(x,y) in the numerator, not just L(a,b). We want to make sure that as we get really close to (but not equal to) (a,b), the function f(x,y) and the tangent plant L(x,y) get close together and become almost indistinguishable from each other. Here's a concrete example showing that your revised definition leads to some issues: Consider the function f(x,y)=x^2+y^2+x+1, which is a polynomial and should be differentiable everywhere. But does your definition of "differentiable" confirm this? We'll show that it says otherwise at (x,y)=(0,0). The equation of the tangent plane to the graph of f at (0,0) is L(x,y)=x+1, and hence L(0,0)=1. Thus, according to your definition, we want to show that lim_{(x,y)->(0,0)} [(x^2+y^2+x+1)-1]/[sqrt(x^2+y^2)]=0. I'll let you verify that this limit actually does not exist, and therefore your definition suggests that f is not differentiable at the origin. If instead we use the definition presented in the video, we must confirm that lim_{(x,y)->(0,0)} [(x^2+y^2+x+1)-(1+x)]/[sqrt(x^2+y^2)]=0, and this is indeed the case.

[@ver-li-o]

​@@mathemation Thanks 😊, I understand. I had that thought because I misunderstood what L(x,y) meant. I know the tangent plane at point (a,b) as f(a,b) + Gf(a,b)·(x-a,y-b), where G is the gradient in this case. In my mind I interpreted L(a,b) as the plane tangent to the graph at point (a,b) and then L(x,y) as the plane tangent to the graph at the point (x,y), and then, as (x,y) moves, the tangential point of the plane to the graph would move too. My L(a,b) was your L(x,y).

[@sohamghosh1865]

hi can you tell me if f(x,y)=xy/|x| (for x not equal to 0 and 0 elsewhere)is differentiable at (0,0) or not, their partial derivatives do exist right?

[@hallgatoandras6511]

The method from 10:40, doesn't apply to every function. For example f(x,y) = ((x^2)*y)/(x^2 + y^2) with this method I get that it is differentiable, but actually it's not. It is continuous in (0,0) but not differentiable. Maybe I missed something.

[@mathemation]

The function you've written isn't defined at (0,0), so it's technically neither continuous nor differentiable there. If you define f(0,0)=0, then f does become continuous at (0,0), and you're correct that f is not differentiable at (0,0). Note that this doesn't contradict anything in the video: all we're claiming is that *if* f fails to be continuous at (a,b), then it also fails to be differentiable at (a,b). As your example shows, there do exist functions that are continuous at (a,b) but may still fail to be differentiable at (a,b). For these functions, you would need to use the definition @5:30 to check differentiability -- this is just a bit beyond what I was expecting of my students in this particular course.

[@hallgatoandras6511]

@@mathemation I see, thank you for the explanation.

[@captainstreamer3261]

By Which Software You Draw Those 3d Structure ??

[@mathemation]

These images were created using Maple. But since making the videos, I have begun using Geogebra for a lot of my 3D graphing. It's freely available here: www.geogebra.org/calculator (just switch to the 3D calculator).

[@glace848]

bro literally speaks in calibri

[@glace848]

good video tho

[@dialgos7574]

This is wrong. A discontinous function CAN be differentiable in R^2. Atleast that proof is an exercise in my class right now

[@mathemation]

If a scalar function is discontinuous at (a,b), then it cannot be differentiable at (a,b). Perhaps you can share what discontinuous function you believe fits the bill and we can discuss.

[@bashydee4936]

How? If it isn't differentiable at each point on its domain, then it isn't. Discontinuity implies it can't be differentiable

[@dialgos7574]

@@mathemation Oh sorry I forgot to answer! The example that we had to examine was (x*y)/(x^2+y^2) which is not continous at (0,0) but it is still differentiable in R^2

[@mathemation]

There are a few things to discuss here. First, as written, this function is not defined at (0,0). You probably actually had some sort of piecewise defined function f with f(0,0)=0 and f(x,y)=xy/(x^2+y^2) otherwise. Second, this function is an interesting example in that although it is not continuous at (0,0), its partial derivatives f_x and f_y exist everywhere -- even at the origin. You can show that f_x(0,0)=0 and f_y(0,0)=0 using the limit definition of the partial derivatives. Perhaps this is what your instructor meant when they said that this function was "differentiable"? As I mentioned in this lesson, we say that a multivariable function is "differentiable" if it can be closely approximated by its tangent plane. This condition is stronger than asking that the partial derivatives exist. Just like in the example 3min into this lesson, the tangent plane to your function at the origin is actually a bad approximation. You can see this by examining the graph of the function and its tangent plane: www.geogebra.org/calculator/m24ppfdp Therefore, this function is *not* differentiable at (0,0).

[@dialgos7574]

@@mathemation Well I guess it comes down to definition. My professor said that the function would be differentiable at (0,0) but we never had the same approach as you. I guess it really doesn't matter too much then :D

[@ed.659]

thank you so much!