# Differentiability of xy function

1. Oct 6, 2009

### gamitor

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

Dear all,

How can I show that the function f(x,y)=xy is differentiable?

Thanks

Dimitris

2. Relevant equations

3. The attempt at a solution

2. Oct 6, 2009

### Dick

What's your definition of 'differentiable' for a function of multiple variables?

3. Oct 6, 2009

### gamitor

I have the definition of differentiability in the pdf file.

View attachment question.pdf

Thanks

4. Oct 6, 2009

### Dick

It can take several hours for an attachment to be approved. Sure you don't want to explain it in words?

5. Oct 6, 2009

### gamitor

The definition is in that link http://en.wikipedia.org/wiki/Derivative

at the 4.4 section

"The total derivative, the total differential and the Jacobian"

The first at 4.4 that appears

Thanks

6. Oct 6, 2009

### Dick

Ok, so in that definition a and h are vectors, yes? Label the components. a=(a_x,a_y), h=(h_x,h_y). Now the formula says if you subtract the linear function f'(a)(h) from f(a+h)-f(a) you are supposed to be left with something that goes to 0 faster than h as h goes to 0. It might help to expand f(a+h)-f(a) in the case where f(x,y)=xy. Do you see a linear part in h?

7. Oct 6, 2009

### HallsofIvy

Staff Emeritus
Well, what you give in the first response is NOT a definition of "differentiable". It certainly is not what is given in the Wikipedia site. I think you must have copied wrong- you have an "f(x,y)" that doesn't belong in there.

Essentially, what Wikipedia says is that f(x,y) is differentiable at $(x_0,y_0)$ if and only if there exist a vector $u\vec{i}+ v\vec{j}$ and a function $\epsilon(x,y)$ such that for for some $f(x,y)= f(x_0,y_0)+ (u\vec{i}+ v\vec{j})\cdot ((x-x_0)\vec{i}+ (y- y_0)\vec{j})+ \epsilon(x,y)$ and that $\lim_{(x,y)\to (x_0,y_0)} \epsilon(x,y)/\sqrt{(x-x_0)^2+ (y-y_0)^2}= 0$.

In that case, the vector $u\vec{i}+ v\vec{j}$ is the "gradient" of f at $(x_0,y_0)$.

So I would recommend that you go ahead and find the gradient of f(x,y)= xy at $(x_0, y_0)$ and use that to determine what $\epsilon(x,y)$ must be. If you can then prove that $\epsilon(x,y)/\sqrt{(x-x_0)^2+ (y- y_0)^2}$ goes to 0 as (x,y) goes to $(x_0,y_0)$, you are done.