Differentiability of a two variable function with parameter

In summary, the question is asking for which value of α the given function is differentiable at the point (0, 0). The attempt at a solution involves evaluating the partial derivatives and taking the limit as (h, k) → (0, 0). The writer also mentions difficulties with two variable limits and requests help. They also acknowledge potential mistakes in their use of English language.
  • #1
Mathitalian
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Homework Statement



For which parameter [tex]\alpha\in\mathbb{R}[/tex] the function:
[tex]f(x, y)= \begin{cases}|x|^\alpha \sin(y),&\mbox{ if } x\ne 0;\\ 0, & \mbox{ if } x=0\end{cases}[/tex]

is differentiable at the point (0, 0)?

The Attempt at a Solution



For α<0, the function is not continue at (0, 0), so it is not differentiable. I checked it :)

For α≥0 i have troubles, many troubles. I need your help.

I started evalueting the partial derivates using the definition:

[tex]\partial_x f(0,0):= \lim_{h\to 0} \frac{|h|^\alpha\sin(0)-0}{h}=0 [/tex]
[tex]\partial_y f(0,0):= \lim_{h\to 0} \frac{|0|^\alpha\sin(h)-0}{h}=0 [/tex]

(I think i must split the cases α=0 and α>0, right? |0|^α has no sense if α=0, but if α=0 then f(x,y)= sin(y)... I'm confuse)

Anyway, i need to find the value of the limit (if it exists):

[tex]\lim_{(h,k)\to (0, 0)}\frac{f(h, k)}{\sqrt{h^2+k^2}}[/tex] and show that its value is 0 right? I have problems with two variables limits...

Any helps will be appreciated...

Please, if you see there are mistakes in English languange, correct me :)
 
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  • #2
See if the limit of each of the following exists as (x, y) → (0, 0):
[itex]\displaystyle \frac{\partial f(x,\,y)}{\partial x}[/itex] and [itex]\displaystyle \frac{\partial f(x,\,y)}{\partial y}[/itex]​
 
Last edited:

1. What is the definition of differentiability for a two variable function with parameter?

Differentiability for a two variable function with parameter means that the function has a well-defined derivative at every point in its domain. This means that the function's rate of change can be calculated at any point, and the function is smooth with no sharp turns or corners.

2. How is differentiability affected by the parameter in a two variable function?

The parameter in a two variable function can affect differentiability in a few ways. If the parameter is a constant, the function may be differentiable at some points but not others. If the parameter is a variable, the function may have differentiability at different values of the parameter, or the function may not be differentiable at all.

3. What is the role of partial derivatives in determining differentiability of a two variable function with parameter?

Partial derivatives play a crucial role in determining differentiability of a two variable function with parameter. If all the partial derivatives of the function exist and are continuous at a certain point, then the function is differentiable at that point. However, if any of the partial derivatives do not exist or are not continuous, then the function is not differentiable at that point.

4. Can a two variable function with parameter be differentiable but not continuous?

Yes, it is possible for a two variable function with parameter to be differentiable but not continuous. This can happen if the partial derivatives exist but are not continuous at a certain point. In this case, the function would have a derivative at that point but would not be continuous.

5. How can the differentiability of a two variable function with parameter be tested?

The differentiability of a two variable function with parameter can be tested using the definition of differentiability, which involves checking the existence and continuity of the partial derivatives at a certain point. Additionally, the differentiability of a function can also be tested by calculating its Jacobian matrix, which is a matrix of all the partial derivatives of the function. If the Jacobian matrix exists and is continuous, then the function is differentiable at that point.

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