Multivariable Calculus | Differentiability

[@ezrasteinberg2016]

Beautiful use of the gradient and the dot product! Woo hoo! Great! Don't stop, Michael!!!! :-)

[@nailabenali7488]

Yeaaaaaay!!! Omg I'm so grateful!! You're really the best !!

[@MichaelPennMath]

I also recorded the proof of differentiable implies continuous and will post later tonight.

[@nailabenali7488]

@@MichaelPennMath I really wanted you to do a video about it but I was just to shy to ask !! Maybe you're not making the world a better place for everyone but you're certainly making it for me and my classmates!! Respect

[@Akash_S_S]

Super class, keep doing more ❤️

[@CDChester]

back at it again with them early uploads!

[@CDChester]

also i'd consider the tree mapping algorithm for the multi-variable and implicit differentiability if you want to intuitively explain it visually

[@henriquenunes7196]

Great example. At min 8,43 the limit is 0,ok, but it should be (x-2), right? Thanks

[@fazioleonardo2821]

I guess it should be : ((x-2)^2)/|x-2|= +/- (x-2)

[@Memes_uploader]

OMG Finally i found such a video

[@keithchan379]

WHY has the criteria of differentiability already assumed f is differentiable by writing fx and fy? Is that not circular logic?

[@jimallysonnevado3973]

This comment may be late but I want to answer your question. Differentiability in higher dimension is less intuitive than differentiability in one variable. Being differentiable in higher dimension only means that you can reasonably estimate a function at a point by using a tangent approximation. However, having partial derivatives, even having all of them, does not automatically mean that a function has a tangent approximation. This is not circular logic because the partial derivatives are defined also in a different way from this (analogous to one dimensional derivative). It just happened that in one variable having a derivative is the same as being differentiable.

[@backyard282]

To check that the function is differentiable at a given point, isn't it enough to check that partial derivatives are defined at the given point? For example in the first example partial with respect to x is 4x and with respect to y is -4, both of which are well defined at (2,-3). Would that be a good proof?

[@duckymomo7935]

no Thm. Existence of partial derivatives does not imply differentiable let f(x, y) = xy/[x^2+y^2] for all (x,y) != (0, 0); 0 if (x, y) = (0, 0) it's just simply not differentiable at (0, 0)

[@michaelempeigne3519]

@@duckymomo7935 What is the name of this theorem that says that?

[@individuoenigmatico1990]

No, the existence alone of partial derivatives is in general not sufficient for differentiability. There are cases where partial derivatives do exist but the function is not differentiable. But the continuity of the partial derivatives at a point gives you the differentiability at that point. That's a theorem. If you have a function f defined on an open ball of a point (a1,...,an) such that the partial derivatives exist in all the points of the open ball and they are all continuous at the point (a1,..,an), then f is differentiable at (a1,...,an). Actually it can be proven that the continuity of all but one partial derivatives in (a1,...,an) and the existence of the last partial derivative in all the open ball is sufficient for differentiability.

[@alieser7770]

I love you sir

[@michaelempeigne3519]

he cancelled the - 4y with the other - 4y in the video but they do not have opposite signs.

[@Sspectator]

Nice haircut.