Homework Help: 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

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.