Totally differentiable?

• springo
In summary, the conversation discussed the differentiability of a function f(x,y) = tan(xy)*sin(1/[x^2+y^2]). It was initially thought that the function was not differentiable because the limit of the function did not exist. However, it was later realized that this was due to a mistake in the calculation. By using polar coordinates and the definition of differentiability, it was determined that the function is in fact differentiable at (0,0).

Homework Statement

For any (x,y) other than (0,0).
$$f(x,y)=\tan(x·y)\sin\left(\frac{1}{x^2+y^2}\right)$$
For (x,y) = (0,0)
f(x,y) = 0
Is f totally differentiable?

The Attempt at a Solution

If the function is not continuous, it can't be differentiable.
$$\lim_{(x,y)\rightarrow(0,0)}\tan(x·y)\sin\left(\frac{1}{x^2+y^2}\right)=\lim_{(x,y)\rightarrow(0,0)}\frac{x·y}{x^2+y^2}=\lim_{r\rightarrow0}\cos(\theta)·sin(\theta)$$
So the limit doesn't exist.

However I looked at the solution and it is differentiable. Is it possible that the solution is wrong or did I make a mistake?

Edit: I don't why it doesn't TeX (at least on my browser) so:
f(x,y) = tan(xy)*sin(1/[x^2+y^2])
And for my attempt at a solution:
lim(x,y->0,0) f(x,y) = lim (x,y->0,0) xy/(x^2+y^2) = lim (r -> 0) cos T*sin T

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Hi springo!

(for some reason the LateX doesn't seem to be working properly today … it's not just you! )

(and try using the X2 tag just above the Reply box )
springo said:
Is f totally differentiable?

If the function is not continuous, it can't be differentiable.

So the limit doesn't exist.

f(x,y) = tan(xy)*sin(1/[x^2+y^2])
And for my attempt at a solution:
lim(x,y->0,0) f(x,y) = lim (x,y->0,0) xy/(x^2+y^2) = lim (r -> 0) cos T*sin T

erm … sin(1/r) is not approximately 1/r, it's ≤ 1,

so |f(x,y)| ≤ tan(xy) -> 0

I was trying to find the limit by using polar coordinates:
x = r·cos(t)
y = r·sin(t)

So lim(x,y)->(0,0) tan(x·y)·sin(1/[x2+y2])
~ lim(x,y)->(0,0) x·y/(x2+y2)
~ limr->0 r2·cos(t)·sin(t)/r2
~ cos(t)·sin(t)
And so the limit doesn't exist.

springo said:
So lim(x,y)->(0,0) tan(x·y)·sin(1/[x2+y2])
~ lim(x,y)->(0,0) x·y/(x2+y2)

Noooo … as r -> 0, sin(r) -> r but sin(1/r) does not -> 1/r, it stays ≤ 1

try drawing it!

OK, I understood! How could I make that mistake...
So how can I do this?

Edit:
I was thinking:
limh->0 [f(x,y) - f(0,0) - fx(0,0)(x-0) - fy(0,0)(y-0)]/√(x2+y2)
Since fx(0,0) = fy(0,0) = 0 (by doing the math... lol) and f(0,0) = 0... it's now:
limh->0 f(x,y)/√(x2+yy) ~ x·y·sin(1/x2+y2)/√(x2+y2)
And by turning into polar we have r2/r = r (the other stuff is bounded so...).
So the limit is 0, so it's differentiable.
Is that right?

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Hi springo!
springo said:
Since fx(0,0) = fy(0,0) = 0 (by doing the math... lol) …

a bit complicated … and sometimes fx(0,0) and fy(0,0) exist when, along some non-axis curve, the derivative doesn't exist.

Just use the definition …

what is lim [f(x,y) - f(0,0)]/r ?