How to prove differentiable everywhere?

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athrun200
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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 [itex]f_{x}[/itex] and [itex]f_{y}[/itex] exist. Obviously, they are differentiable for x and y[itex]\neq[/itex]0

For part c.
It seems there are 2 method to do it.
1. Use first principle.(i.e. take limit)
2.Find [itex]f_{xy}[/itex] and [itex]f_{yx}[/itex]
If they equal each other, then f is differentiable at(0,0)
 
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  • Printed math problem 3 defining f(x, y) piecewise: (2x^3 - x^2 y)/(x^2 + y^2) for (x, y) not equal to (0, 0), and 0 at (0, 0); parts (a), (b) and (c) ask about partial derivatives and differentiability.
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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 [itex]f_{x}[/itex] and [itex]f_{y}[/itex] exist. Obviously, they are differentiable for x and y[itex]\neq[/itex]0

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 [itex]f_{xy}[/itex] and [itex]f_{yx}[/itex]
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.
 
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?

Limit as h approaches 0 of [f(x0 + h) - f(x0) - J(x0)h] divided by the norm of h, equal to 0.
 
I also wonder if [itex]f_{x}[/itex]=[itex]\frac{(2^xy^3-6x^2y^2-2x^4)}{(y^4+2x^2y^2+x^4)}[/itex] exist at (0,0).

If not, it seems the question part a has some problems.