# Differentiability of a multi-variable function

1. Oct 3, 2012

### dumbQuestion

For a function of a single variable, I can check if the function is differentiable by simply taking the limit definition of a derivative and if the limit exists, then the function is differentiable at that point. Differentiability also implies continuity at this level.

Now, for a function of say, f(x,y), a function of just two variables, if I want to know if f is differentiable at a point c things seem more complicated. I know I can take the limit definition of partial derivatives and say check if the limit for the partial with respect to x exists at c, and if the limit for the partial with respect to y exists at c. But it seems this won't be sufficient because I know that there are infinitely many directional derivatives. Wouldn't all of those directional derivatives need to exist at c for the function to be differentiable at c? Is there some general way to check this withi a limit definition, say by looking at the gradient? (I am asking about the gradient because I know I can find a directional derivative in any direction u by taking the dot product of the gradient at c and the unit vector in the direction of u)

Also, if every single partial derivative exists, can we then say the function is continuous as a result of being differentiable?

(A side question: what if on the single variable level I have a piecewise function like:

f(x) = {(x^2-4x+4)/(x-2) if x =/= 2,
0 if x x==2}

This function would be differentiable at x = 2 right? I'm sorry but the piecewise functions just confuse me when it comes to these things)

2. Oct 3, 2012

### arildno

1. "But it seems this won't be sufficient because I know that there are infinitely many directional derivatives. Wouldn't all of those directional derivatives need to exist at c for the function to be differentiable at c? "
Absolutely! You are very perceptive, it is a subtle, but important point!
2. "Is there some general way to check this withi a limit definition, say by looking at the gradient? "
Yes. For multivariable functions, it is not considered differentiable unless you can prove that limit holds for all points within every sequence of shrinking REGIONS about the point. In effect, this will hold if you can show it holds for every sequence of shrinking BALLS around the point, since for every particular choice of a sequence of shrinking generel regions you might make, ypu'll be able to find some tiny ball contained within such regions (remember that open sets are defined in that for every point within the set, you'll be able to create a tiny ball around out, wholly enclosed in the region!)

Thus, utilizing polar coordinates is a critical tool for many evaluations of limits in the multi-variable case; and the requirement then is that the limit you get must be independent of any of the angular variables as the "radius" goes to zero.

3."Also, if every single partial derivative exists, can we then say the function is continuous as a result of being differentiable? 2
All directional derivatives of a function may exist, but it might STILL be..undifferentiable. Nasty, but true.

4. The simplest way to draw an analogue to the 1-dimensional case for the multivariable functions is to remember that partial derivatives only evaluates the fraction along a strict SUBSET of all numbers floating about your point (say, those along the x-axis, or those points along the y-axis)
What is similar to this for the one-variable case?
Simply, DIFFERENT sequences converging to zero (for example, along the irrationals or along the rationals)

Just because the function CAN be "differentiated" when we restrict our attention to the numbers contained in that sequence, does not mean the function itself is differentiable.
The limit must hold for ALL choices of converging sequences, not just one of them.

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