20060529, 13:41  #1 
May 2005
Argentina
2×3×31 Posts 
Multivariable differentiability
Sorry if this is a bit offtopic.
I searched the web for the proof that if a multivariable function has partial derivatives, and they are continuous, then it is differentiable. (and f is called a C1 function) Any help will be appreciated. Damian. 
20060530, 04:06  #2 
Cranksta Rap Ayatollah
Jul 2003
641 Posts 
p. 219 of Rudin, Theorem 9.21

20060530, 12:53  #3 
May 2005
Argentina
2·3·31 Posts 
Thanks for the help. I don't have Rudin's book but neitherway I found the proof here http://www.maths.mq.edu.au/~wchen/ln...er/mva02d.pdf
Now I have another question related to the same theorem. Suppose the function: f(x,y) ={ 1 (if x=0 or y=0) { 0 (otherwise) the partial derivatives at the origin are df/dx = 0 and df/dy = 0 so they exist and are continious. However the function is not continious in (0,0), so it can't be differentiable there. (there can't be a tangent plane) What did I misundertood? Because I think the theorem says that if the partial derivatives exists and they are continious then the function is differentiable there. 
20060530, 18:26  #4  
∂^{2}ω=0
Sep 2002
República de California
2×3×1,931 Posts 
Quote:
Also, it makes no sense to speak of a function being continuous at a single point  continuity only makes sense in regions, i.e. neighborhoods of a point. 

20060530, 19:03  #5  
"William"
May 2003
New Haven
2^{3}×5×59 Posts 
Quote:
f(x) = 1/q if x is rational p/q in lowest form, f(x)= 0 if x is irrational This function is continuous at irrational points. (f(x) is continuous at x' if, for every epsilon > 0 there exists a delta such that for every x such that xx'<delta, f(x)f(x')<epsilon) William 

20060601, 13:15  #6 
May 2005
Argentina
272_{8} Posts 
Thanks for the help. I've been thinking about it and I think I understand the flaw in the statement I posted above now.
The theorem says the functions partial derivatives have to be continious in the neighborhood of a point, in order to be diferentiable there, wich isn't the case in the f(x,y) I formulated above in the origin, so it does't say nothing about this function been differentiable in the origin. Damian. 
20060601, 13:20  #7 
May 2005
Argentina
2×3×31 Posts 
Another question: Does anybody knows who was the mathemathician that first proved that theorem? It sounds strange to me that the theorem hasn't got a name asociated with it.

20060602, 13:40  #8 
May 2005
Argentina
BA_{16} Posts 
Has anybody one example of a function that is differentiable at a point, but is not a class C1 function at that point?
Is that even possible? Thanks, Damian. 
20060602, 17:29  #9  
∂^{2}ω=0
Sep 2002
República de California
2·3·1,931 Posts 
Quote:
There are eminently readable online links to all of this stuff, you know  for example: http://mathworld.wolfram.com/Derivative.html http://mathworld.wolfram.com/CkFunction.html 

20060602, 18:45  #10 
May 2005
Argentina
2×3×31 Posts 
Thanks for your reply.
The {Since "C1" is equivalent to "derivative exists"} was what I needed to know So what do you think, is it ok to say: a function is differentiable in the neighborhood of a point A, if and only if its partial derivatives are continious in the neighborhood of A? In other words, is it a neccesary and sufficient condition for a function to have partial derivatives continious in an interval, to be differentiable there? 
20060602, 19:28  #11 
∂^{2}ω=0
Sep 2002
República de California
26502_{8} Posts 
kdifferentiable (k = 0,1,2,...) in an interval/region means derivatives (or partial derivatives, if we're in 2 or more dimensions) of order <= k exist in that interval/region.

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