Differentiability of xy function

[gamitor]

Homework Statement

Dear all,

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

Thanks

Dimitris

Homework Equations

The Attempt at a Solution

 

[Dick]

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

 

[gamitor]

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

I have the definition of differentiability in the pdf file.

View attachment question.pdf

Thanks

 

[Dick]

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

 

[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

 

[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?

 

[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.