Partial derivative

  • Thread starter Telemachus
  • Start date
  • #1
Telemachus
835
30

Homework Statement


Well, I'm not sure about this one. Its actually a differentiation problem, it asks me to determine if the function is differentiable at the given point.

[tex]\sqrt[ ]{|xy|}[/tex] at [tex]P(0,0)[/tex]
I think its not, but I must demonstrate, off course.

So I try to solve the partial derivatives:

[tex]f_x=\begin{Bmatrix} \displaystyle\frac{|y|}{2\sqrt[ ]{|xy|}} & \mbox{ if }& x>0\\\displaystyle\frac{-|y|}{2\sqrt[ ]{|xy|}} & \mbox{if}& x<0\end{matrix}[/tex]
And here is the deal. Is this right? if it is, its easy to see that the partial derivative is not continuous at (0,0), its actually not defined at that point. So its not differentiable.

Bye there, and thanks.
 

Answers and Replies

  • #2
36,477
8,439
This is similar to f(x) = sqrt(|x|), which is defined and continuous for all real x. The graph has a cusp at (0, 0), so the derivative is not defined there.
 
  • #3
Telemachus
835
30
Thanks Mark. So what I did is ok, right?
 
  • #4
36,477
8,439
I think you have a mistake because of the absolute value.
For example, if
[tex]f(x) = \sqrt{|x|} = |x|^{1/2}[/tex]

then
[tex]f'(x) = (1/2) |x|^{-1/2} \cdot d/dx \left(|x|\right)[/tex]

In the above, d/dx|x| = |x|/x.

This derivative works out to sqrt(|x|)/(2x).

In your partial, it looks like you're missing the step where you take the partial with respect to x of |xy|.
 
  • #5
Telemachus
835
30
Thanks.
So this:
[tex]f'(x) = (1/2) |x|^{-1/2} \cdot d/dx \left(|x|\right)[/tex]

Its equal to this:

[tex]f'(x)=\begin{Bmatrix} \displaystyle\frac{1}{2\sqrt[ ]{|x|}} & \mbox{ if }& x>0\\\displaystyle\frac{-1}{2\sqrt[ ]{|x|}} & \mbox{if}& x<0\end{matrix}[/tex]

Isn't it?
 
  • #7
Telemachus
835
30
So, if we take [tex]\sqrt[ ]{|xy|}[/tex]

Then
[tex]f_x=\begin{Bmatrix} \displaystyle\frac{|y|}{2\sqrt[ ]{|xy|}} & \mbox{ if }& x>0\\\displaystyle\frac{-|y|}{2\sqrt[ ]{|xy|}} & \mbox{if}& x<0\end{matrix}[/tex]
Cause in the numerator I've got the derivative of |x| multiplied by |y|, which is a constant.

I see the mistake now. I am considering just when x>0 and x<0, and not xy>0 and xy<0.
 
Last edited:

Suggested for: Partial derivative

Replies
3
Views
406
  • Last Post
Replies
5
Views
386
Replies
7
Views
351
  • Last Post
Replies
1
Views
526
Replies
5
Views
722
Replies
22
Views
925
Replies
3
Views
342
Replies
10
Views
442
Replies
8
Views
409
Replies
0
Views
188
Top