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Totally differentiable?

  • Thread starter springo
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  • #1
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Homework Statement


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

Homework Equations



The Attempt at a Solution


If the function is not continuous, it can't be differentiable.
[tex]\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)[/tex]
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?

Thanks for your help.

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
 
Last edited:

Answers and Replies

  • #2
tiny-tim
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Hi springo! :smile:

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

(and try using the X2 tag just above the Reply box :wink:)
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 :redface: … sin(1/r) is not approximately 1/r, it's ≤ 1,

so |f(x,y)| ≤ tan(xy) -> 0 :wink:
 
  • #3
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Hi tiny-tim! Thanks for your answer.
I'm not sure I understand what you meant with your answer.
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.
 
  • #4
tiny-tim
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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! :smile:
 
  • #5
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OK, I understood! How could I make that mistake... :rolleyes:
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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  • #6
tiny-tim
Science Advisor
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Hi springo! :smile:
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 ? :smile:
 

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