Partial Derivative Homework: Is \sqrt[ ]{|xy|} Differentiable at (0,0)?

In summary, the conversation discusses a differentiation problem that asks if a function is differentiable at a given point. The participants try to solve the partial derivatives and determine that the function is not differentiable due to a cusp at the given point. They also discuss a mistake made in considering only positive and negative values of x instead of xy.
  • #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.
 
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  • #2
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
Thanks Mark. So what I did is ok, right?
 
  • #4
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
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
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:

Related to Partial Derivative Homework: Is \sqrt[ ]{|xy|} Differentiable at (0,0)?

1. What is a partial derivative?

A partial derivative is a mathematical concept used in multivariable calculus to calculate the rate of change of a function with respect to one of its independent variables, while holding all other variables constant.

2. How is a partial derivative calculated?

To calculate a partial derivative, you take the derivative of the function with respect to the chosen independent variable, treating all other variables as constants. This is denoted by ∂ (the partial derivative symbol) followed by the variable with respect to which the derivative is being taken.

3. What does it mean for a function to be differentiable?

A function is differentiable if it has a well-defined derivative at every point in its domain. In other words, the function must be smooth and continuous with no sharp turns or breaks.

4. Is the function √|xy| differentiable at (0,0)?

No, the function √|xy| is not differentiable at (0,0) because it is not continuous at that point. The function has a sharp turn at (0,0) which means it does not have a well-defined derivative at that point.

5. Can a function be partially differentiable but not differentiable?

Yes, a function can be partially differentiable but not differentiable. This means that the function has well-defined partial derivatives with respect to each of its independent variables, but it is not differentiable at certain points due to discontinuities or sharp turns.

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