Determining domain for C^1 function

lys04
Messages
144
Reaction score
5
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
 
Last edited by a moderator:
Physics news on Phys.org
I don’t think latex is working, not sure what’s wrong with it sorry
 
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.
 
Works very well with the additional }:

$$L \frac{4x}{5(x^{2}+y^{2})^{\frac{-3}{5}}}$$
 
\sqrt[2n+1]{x} is not differentiable at x = 0 for n \geq 1. This follows from the fact that x^{2n+1} has a point of inflection here.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top