Proving a function of 2 variables is diffable

In summary, the problem is to prove the differentiability of the function f(x,y) = (x^4 + y^4)/((x^2+y^2)^alpha) on R^2 for alpha < 3/2. This can be done by showing that the partial derivatives of f are continuous for all points except (0,0), and then using the definition of differentiability. However, it is unclear where the condition alpha < 3/2 comes into play.
  • #1
SNOOTCHIEBOOCHEE
145
0

Homework Statement



i)Prove that f: [tex]R^2\rightarrow R[/tex]

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

is differentiable on [tex]R^2[/tex] for [tex]\alpha[/tex]<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: [tex]R\rightarrow R^m[/tex] and there is a T (element) L(R^n;R^m) such that the function
[tex]\epsilon[/tex](h) := f(a+h)-f(a)-T(h) satisfies [tex]\epsilon[/tex](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!
 
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  • #2
Bump...

i still need help on this problem
 
  • #3
Well, suppose that [itex]\alpha<3/2[/itex], 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
I don't know if this method would work.

fx = [tex]\frac{(x^2 + y^2)4x^3 - 2ax(x^4+y^4)}{(x^2+y^2)^a+^1}[/tex]And this is continuous for all (x,y)==/==0

but i don't see where 3/2 falls out of there.
 
Last edited:

Related to Proving a function of 2 variables is diffable

1. What does it mean for a function of 2 variables to be diffable?

When a function of 2 variables is diffable, it means that it is differentiable at every point in its domain. This means that the function has a well-defined tangent plane at every point in its domain, and its partial derivatives exist at every point.

2. How can I prove that a function of 2 variables is diffable?

In order to prove that a function of 2 variables is diffable, you need to show that its partial derivatives exist at every point in its domain and that they are continuous. This can be done using the limit definition of a derivative and the continuity test for multivariable functions.

3. What is the significance of a function of 2 variables being diffable?

A function of 2 variables being diffable is significant because it means that the function is smooth and well-behaved at every point in its domain. This allows for the use of calculus techniques, such as finding the slope of a tangent line or calculating the rate of change, to analyze the behavior of the function.

4. Can a function of 2 variables be diffable at some points but not others?

Yes, a function of 2 variables can be diffable at some points but not others. This means that the function is not diffable at points where its partial derivatives do not exist or are not continuous. It is important to consider the entire domain of a function when determining its diffability.

5. Are there any special cases when proving a function of 2 variables is diffable?

Yes, there are a few special cases to consider when proving a function of 2 variables is diffable. These include functions with sharp corners or discontinuities in their domain, as well as functions that are not defined at certain points. In these cases, additional techniques may be needed to prove the function's diffability.

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