Partial derivative exists at origin but not continuous there

[mathFun]

I always see the example


f(x,y)={xy/(x²+y²) if (x,y) =/= (0,0) and 0 if (x,y)=(0,0)}


given as the example of a function where the partial derivatives exist at the origin but are not continuous there. I have a difficult time wrapping my head around this and was hoping someone could check my logic.



Now let's consider the partial derivatives f_x and f_y


f_x = [3x²y + y³]/(x²+y²)² if (x,y)=/=(0,0) and 0 if (x,y) = (0,0)


Defining it piecewise like this means f_x is defined at the origin, but its not continuous at the origin because of the same problem: it isn't continuous because say we approach the origin from the x axis, then this is 0, but if we approach it from the y axis, then the limit doesn't exist because you get lim x-->(0,0) 1/y = ∞. So its not continuous here because the limits don't match.


One problem is I have a difficult time understanding why the partials exist though. Because the original function f(x,y) isn't continuous at the origin. For example, if you approach the origin from the parabola y=x², you get lim (x,y)--> (0,0) [x³/(x²+x⁴)] (we end up using L'Hopitals Rule since this is 0/0 indeterminate, and eventually end up with the limit going to infinity because the largest power is in the denominator). So it doesn't seem like the limit of the original function exists at the origin, which would mean its not continuous at the origin, which would mean the derivatives don't exist at the origin right? Because being continuous is essential to being differentiable? Is being continuous essential to being partial differentiable?


I'm so confused!

 

[micromass]

You seem to think that if the partial derivatives exist at a point, that the function must be continuous there. This is not true. And your example is a counterexample for that.

We do have the following results:

• If a function is differentiable at a point, then it is continuous.
• If the partial derivatives exists at a point and are continuous there, then the function is differentiable at that point and the function is continuous there.

With differentiable, I mean that the Frechet derivative exists: http://en.wikipedia.org/wiki/Fréchet_derivative (although multivariable calculus courses use the term Frechet derivatives, but will simply speak of differentiable functions).

 

[dumbQuestion]

I think I am finally seeing why I was so confused before. I kept thinking "partial derivative exists at c" and that sounded so strong to me, as though the function were differentiable at c. But the analogue to 2D of a partial derivative existing is more like saying the one sided limit exists, since a partial derivative is just a directional derivative. It doesn't give the whole picture of what's going on.So just like in 2D, in order for a function to be differentiable at a point c, both sides of the limit of the difference equation have to exist and be equal. Similarly, I am assuming in 3D, every direction derivative (including f_x and f_y then) would have to exist and be equal. Is that more along the lines?

 

[micromass]

I think I am finally seeing why I was so confused before. I kept thinking "partial derivative exists at c" and that sounded so strong to me, as though the function were differentiable at c. But the analogue to 2D of a partial derivative existing is more like saying the one sided limit exists, since a partial derivative is just a directional derivative. It doesn't give the whole picture of what's going on.


So just like in 2D, in order for a function to be differentiable at a point c, both sides of the limit of the difference equation have to exist and be equal. Similarly, I am assuming in 3D, every direction derivative (including f_x and f_y then) would have to exist and be equal. Is that more along the lines?

I'm afraid not, things in 3D get very weird. The following link contains an example of a discontinuous functions whose directional derivatives exists and are equal.

http://calculus.subwiki.org/wiki/Ex...in_every_direction_not_implies_differentiable

 

[mathFun]

hmm... I guess I got to just keep thinking about this and eventually it will make sense. :/