Proving a function of 2 variables is diffable

[SNOOTCHIEBOOCHEE]

Homework Statement

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

Homework 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

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. please HALP MEH!

 

[SNOOTCHIEBOOCHEE]

Bump...

i still need help on this problem

 

[morphism]

Well, suppose that \alpha&lt;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.]

 

[SNOOTCHIEBOOCHEE]

I don't know 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 don't see where 3/2 falls out of there.