# Proving a function of 2 variables is diffable

1. Feb 7, 2008

### SNOOTCHIEBOOCHEE

1. The problem statement, all variables and given/known data

i)Prove that f: $$R^2\rightarrow R$$

f(x,y)= $$\frac{x^4 + y^4}{(x^2+y^2)^\alpha}$$ (x,y) =/= (0,0)
f(x,y)= 0 (x,y)=(0,0)

is differentiable on $$R^2$$ for $$\alpha$$<3/2

2. Relevant equations

let f be a vector function from n variable sto m variables
i) f is said to be diffable at a point a (element) R^n iff there is an open set V containing a such that f: $$R\rightarrow R^m$$ and there is a T (element) L(R^n;R^m) such that the function
$$\epsilon$$(h) := f(a+h)-f(a)-T(h) satisfies $$\epsilon$$(h)/||h|| ---> 0 as h--->0

ii) f is said to be differentiable on a set E iff E is not empty and f is diffable at every point in E

3. The attempt at a solution

I actually have no clue how to do this problem. I had my wisdom teeth pulled the day he covered this in lecture. plz HALP MEH!!!

2. Feb 10, 2008

### SNOOTCHIEBOOCHEE

Bump...

i still need help on this problem

3. Feb 10, 2008

### morphism

Well, suppose that $\alpha<3/2$, and consider the partials of f(x,y) and see if they are continuous. If they are, then f is differentiable.

[Note: I haven't tried this.]

4. Feb 10, 2008

### SNOOTCHIEBOOCHEE

I dunno if this method would work.

fx = $$\frac{(x^2 + y^2)4x^3 - 2ax(x^4+y^4)}{(x^2+y^2)^a+^1}$$

And this is continuous for all (x,y)==/==0

but i dont see where 3/2 falls out of there.

Last edited: Feb 10, 2008
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