Real Analysis (vector cal proof): If a partial derivative is discontinuous at a point?

Question

can I then say that the function is not differentiable at that point?

I'm supposed to show whether or not
f ((x,y)) = √(|xy|)
is differentiable at (0,0).

I trying to prove that it is not differentiable at (0,0). I wanted to say that since df/dx is not continuous is not continuous at (0,0) then the function is not differentiable. Is this right?

Let me rephrase: I wanted to say that since lim (x,y) -> (0,0) df/dx DNE
then the function is not differentiable. Is this right?

I cannot just say that it is obvious that the absolute value function is not differentiable at (0,0). I would need to prove it.

Answer 1

That's not the case. Consider the basic 1-variable function f(x) = {x² sin (1/x) if x≠0, 0 if x=0}. This function may be seen to be differentiable everywhere as follows:

If x≠0, then using the standard rules of differentiation, we see that f'(x) = 2x sin (1/x) - cos (1/x)

If x=0, then going to the definition of the derivative, we see that:

f'(0) = [h→0]lim (f(h) - f(0))/h
= [h→0]lim (h² sin (1/h) - 0)/h
= [h→0]lim h sin (1/h)
= 0

However, while f is everywhere differentiable, the derivative is not continuous -- indeed, [x→0]lim f'(x) = [x→0]lim 2x sin (1/x) - cos (1/x) doesn't even exist, and thus a fortiori cannot be equal to f'(0).

The above example may be easily extended to yield a two-variable function whose partial derivatives are not continuous -- e.g. let f(x, y) = x² sin (1/x) -- then ∂f/∂x exists everywhere, but is discontinuous at (0, y) for all y (of course, ∂f/∂y is simply the constant function 0).

Now, going back to the particular example of √|xy|, the simplest way to show that this function is not differentiable is to note that if a curve is differentiable at a point, then the composition of that function with any differentiable curve through the point will also be differentiable at that point (by the chain rule). So if you can exhibit a differentiable curve γ(t) such that γ(t) is differentiable at t but f(γ(t)) is not differentiable at t, it will follow that f(γ(t)) cannot possibly be differentiable at γ(t). In this case, the curve γ(t) = (t, t) will work just fine. Then we have:

f(γ(t)) = √|tt| = √t² = |t|, which is not differentiable at zero, so √|xy| cannot be differentiable at γ(0) = (0, 0).

Edit: "Let me rephrase: I wanted to say that since lim (x,y) -> (0,0) df/dx DNE
then the function is not differentiable. Is this right?"

And my point is NO, this is not right, it is quite possible for [(x, y)→(0, 0)]lim (∂f/∂x)(x, y) not to exist, but for (∂f/∂x)(0, 0) to exist (and in fact I gave you an explicit example of such a function).

What I'm suggesting to you is that instead of working with the two variable function, you choose a suitable differentiable curve γ passing through (0, 0) and argue that if f(x, y) is differentiable, so is f(γ(t)). Conversely, if f(γ(t)) is not differentiable, then neither is f(x, y), so this reduces the problem of proving a two-variable function has no derivative at a point to proving that a particular one-variable function has no derivative at a point, which is much easier. In this case, using the curve γ(t) = (t, t), we have that f(γ(t)) = |t|, and it's relatively easy to prove that |t| is not differentiable at 0 (you should have done it several times in univariate calculus).

Answer 2

If the partial derivative is discontinous is it defined at the point? Probably not unless is it given a value in the question. Then you can say its not differentiable at that point.

Counter-example. The dirac delta function is integrable under certain conditions to get a constant function. This is why I ask if the point on the derivative is defined.