Determining domain for C^1 function

  • #1
lys04
70
3
Homework Statement
Picture
Relevant Equations
Partial derivatives
The ####x partial derivative is equal to $$L \frac{4x}{5(x^{2}+y^{2})^{\frac{-3}{5}}}$$ and the partial for ##y## is $$L \frac{4y}{5(x^{2}+y^{2})^{\frac{-3}{5}}}$$
Using the limit definition of partial derivatives I got the partial wrt ##x## is $$L \frac{h^{\frac{4}{5}}}{h}$$ which doesn’t exist as ##h## goes to ##0##. Similar argument for partial wrt ##y##. This means that ##f## isn’t ##C^1## at the origin, right?

At every other point the partial derivatives exist and is continuous because it’s a composition of a polynomial of two variables and ##x^2/5##, so ##f## is ##C^1## at all points except the origin.

Is the reasoning correct?
IMG_0425.jpeg
 
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  • #2
I don’t think latex is working, not sure what’s wrong with it sorry
 
  • #3
lys04 said:
I don’t think latex is working, not sure what’s wrong with it sorry
One problem is that { and } don't match.
 
  • #4
Works very well with the additional }:

$$L \frac{4x}{5(x^{2}+y^{2})^{\frac{-3}{5}}}$$
 
  • #5
[itex]\sqrt[2n+1]{x}[/itex] is not differentiable at [itex]x = 0[/itex] for [itex]n \geq 1[/itex]. This follows from the fact that [itex]x^{2n+1}[/itex] has a point of inflection here.
 

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