Partial derivative

1. Sep 21, 2010

Telemachus

1. The problem statement, all variables and given/known data
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.

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

So I try to solve the partial derivatives:

$$f_x=\begin{Bmatrix} \displaystyle\frac{|y|}{2\sqrt[ ]{|xy|}} & \mbox{ if }& x>0\\\displaystyle\frac{-|y|}{2\sqrt[ ]{|xy|}} & \mbox{if}& x<0\end{matrix}$$
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.

2. Sep 21, 2010

Staff: Mentor

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. Sep 21, 2010

Telemachus

Thanks Mark. So what I did is ok, right?

4. Sep 21, 2010

Staff: Mentor

I think you have a mistake because of the absolute value.
For example, if
$$f(x) = \sqrt{|x|} = |x|^{1/2}$$

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

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. Sep 21, 2010

Telemachus

Thanks.
So this:
$$f'(x) = (1/2) |x|^{-1/2} \cdot d/dx \left(|x|\right)$$

Its equal to this:

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

Isn't it?

6. Sep 21, 2010

Staff: Mentor

Yes.

7. Sep 21, 2010

Telemachus

So, if we take $$\sqrt[ ]{|xy|}$$

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

I see the mistake now. Im considering just when x>0 and x<0, and not xy>0 and xy<0.

Last edited: Sep 21, 2010