# Thread: Partial derivative, differentiation problem

1. ## Partial derivative, differentiation problem

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 for posting.

2. Originally Posted by Ulysses
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 for posting.
There's a difference between "the partial derivatives at (0,0) exist but aren't continuous" and "the partial derivatives at (0,0) don't exist": in the former case the function may be differentiable at (0,0), but in the latter it can't, since differentiability in a point ==> existence of partial derivatives at that point.

Tonio

3. Ok, you're right, but what I meant was that the partial derivative wasn't continuous at that point, so then the tangent plane doesn't exists.

Here, from the Stewart:

Theorem If the partial derivatives $f_x$ and $f_y$ exist near $(a,b)$ and are continuous
at $(a,b)$, then is differentiable at $(a,b)$.

I have doubts about the derivative it self, I think that my solution is wrong.