# Partial derivative

## 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.

$$\sqrt[ ]{|xy|}$$ 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.

Mark44
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.

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

Mark44
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|.

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?

Mark44
Mentor
Yes.

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.

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