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I'm trying to understand differentiability of multivariable functions.

The textbook says, "If the partial derivatives ƒx and ƒy of a function ƒ(x, y) are continuous throughout an open region R, then ƒ is differentiable at every point of R."

Hass, Joel R.; Heil, Christopher E.; Weir, Maurice D.. Thomas' Calculus (Page 818). Pearson Education. Kindle Edition.

So in two dimensions, if something is continuous, it might not be differentiable, because it could be pointy (that's an official math term, right?) Couldn't that happen in three dimensions too?

Also, I was wondering whether the converse of the above is true - i.e. if a multivariable function is differentiable, that means it's continuous and that the partial derivatives exist. And if not, what's the counterexample?

Thank you!

J. W. Tanner
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user754601
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2 Answers2

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No, “pointy” is not an official math term. Not only there is no such thing as official math terms, as I have never seen it on a mathematical text.

But, yes, just like in the case of functions from $\mathbb R$ into $\mathbb R$, a continuous functions may fail to be differentiable. An example would be$$\begin{array}{ccc}\mathbb R&\longrightarrow&\mathbb R\\x&\mapsto&\begin{cases}x\sin\left(\frac1x\right)&\text{ if }x\neq0\\0&\text{ otherwise.}\end{cases}\end{array}$$By the way, the graph of this function is not “pointy”.

And, yes, this can also happens in the context of function from $\mathbb R^n$ into $\mathbb R$.

On the other hand, asserting that a function $f$ is differentiable does not mean that $f$ is continuous and that the partial derivatives exists. It is stronger than that. An example would be$$\begin{array}{rccc}f\colon&\mathbb R^2&\longrightarrow&\mathbb R\\&(x,y)&\mapsto&\begin{cases}\frac{xy}{x^2+y^2}&\text{ if }(x,y)\neq(0,0)\\0&\text{ otherwise.}\end{cases}\end{array}$$It has partial derivatives everywhere, but it is not differentiable at $(0,0)$.

José Carlos Santos
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  • I am sorry but your last paragraph seems a bit off to me. Because differentiability does imply continuity and also the existence of the partial derivatives - they do not need to be continuos though. And then your example also doesn't claim what your said before. But it shows that only the existence of partial derivatives does not imply differentiability. – hal4math Feb 26 '20 at 18:23
  • I suppose that there is a language problem here. I wrote what I wrote assuming that when someone says that $A$ means $B$, then that person is asserting that $A\iff B$. Your comment seems to assume that that person is just asserting that $A\implies B$. – José Carlos Santos Feb 26 '20 at 18:27
  • Ah, okay, I am sorry. Thank you for the clarification. I think I indeed didn't read the "means" to carefully because I understood OPs question rather as an "implies", or more precisely, I was more concerned with the " - i.e " part of the Question and not with the correct converse statement which you provided an answer to. – hal4math Feb 26 '20 at 18:37
  • Did you mean $f:\mathbb R^\color{red}2\to\mathbb R$? – J. W. Tanner Feb 26 '20 at 19:11
  • Thanks for your help on this. By "pointy," I meant like the absolute value graph. I'm imagining a mountain that if you go over it in the x direction or in the y direction, it is rounded, concave down. But wouldn't it still be possible in between there, say if you walk in the i+j direction for the mountain to be just slanted straight up to a point and then straight down? And would it still be differentiable at that top point? – user754601 Feb 26 '20 at 19:46
  • In the other part of my question, I was trying to ask whether knowing that a multivariable function is differentiable at a particular point means it has to be continuous at that point and whether it has to have partial derivatives at that point. Sorry I wasn't more clear. – user754601 Feb 26 '20 at 19:49
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    @J.W.Tanner Yes. I've edited my answer. Thank you. – José Carlos Santos Feb 26 '20 at 21:15
  • @user754601 If $f$ is differentiable at a point, then it is continuous there and it has partial derivatives at that point. But being differentiable is stronger than this. – José Carlos Santos Feb 26 '20 at 21:17
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Note that the book does not state that $f$ needs to be continuous, but rather that the partial derivatives are continuous. So you actually misinterpreted what the converse of the theorem would be.

The converse would be that, if $f$ is differentiable, then the partial derivatives exists and are continuous.

The first part of the statement is true, the partial derivatives do exist if $f$ is differentiable, but the partial derivatives are not necessarily continuous, not even in the one dimensional case. The classic counter example would be $$x \mapsto \begin{cases} x^2 sin(\frac{1}{x}) &, x\neq 0 \\ 0 & ,x=0 \end{cases}$$ For more examples see Discontinuous derivative.

Leander Tilsted Kristensen
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