How to prove differentiable everywhere?

  • Thread starter Thread starter athrun200
  • Start date Start date
  • Tags Tags
    Differentiable
Click For Summary
To prove a function is differentiable everywhere, it is essential to establish the existence of partial derivatives and their continuity. For part b, the function is differentiable for x and y not equal to zero, as it is a quotient of polynomials. In part c, two methods are suggested: using the first principle of limits and checking if the mixed partial derivatives are equal. However, it is clarified that differentiability is not solely defined by limits, and the converse of the mixed derivatives theorem does not hold. The discussion raises concerns about the existence of derivatives at the origin and potential issues in part a of the problem.
athrun200
Messages
275
Reaction score
0

Homework Statement


See photo, part b and c


Homework Equations





The Attempt at a Solution


For part b
It seems it is trival, in part a we have proved that f_{x} and f_{y} exist. Obviously, they are differentiable for x and y\neq0

For part c.
It seems there are 2 method to do it.
1. Use first principle.(i.e. take limit)
2.Find f_{xy} and f_{yx}
If they equal each other, then f is differentiable at(0,0)
 

Attachments

  • 擷取.PNG
    擷取.PNG
    18.6 KB · Views: 1,174
Physics news on Phys.org
athrun200 said:

Homework Statement


See photo, part b and c


Homework Equations





The Attempt at a Solution


For part b
It seems it is trival, in part a we have proved that f_{x} and f_{y} exist. Obviously, they are differentiable for x and y\neq0

For part c.
It seems there are 2 method to do it.
1. Use first principle.(i.e. take limit)
What limit are you talking about? "Differentiability" of a function of two variables is NOT defined by a limit. You might want to look up the definition of "differentiable" for functions of more than one variable.

2.Find f_{xy} and f_{yx}
If they equal each other, then f is differentiable at(0,0)
No, that's not true. There is a theorem that says "If a function is differentiable, on a region, then its mixed second derivatives are equal on that region", but the converse of that statement is not true.
 
How about part b?
You only talk about part c
 
Yes, (b) is trivial- f is a quotient of two polynomials and the denominator is not 0 for any point other than (0, 0).
 
HallsofIvy said:
What limit are you talking about? "Differentiability" of a function of two variables is NOT defined by a limit. You might want to look up the definition of "differentiable" for functions of more than one variable.
But I saw from wiki that it is.
If not, how do I prove differentiable?

cb2be0dc4607423c38120e364c9d4a65.png
 
I also wonder if f_{x}=\frac{(2^xy^3-6x^2y^2-2x^4)}{(y^4+2x^2y^2+x^4)} exist at (0,0).

If not, it seems the question part a has some problems.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
3
Views
1K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
6
Views
2K
Replies
1
Views
2K
Replies
14
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K