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derivative

explain why this function is differentiable at every point in its domain

f(x,y)=((xy^2)/(x^2+y^4),x/y+y/x)

Western, I think you have a typo in your question. What does the comman (,) mean? Should that be a + or - or something?

Anyway, if you compute $f_x(x,y)=\frac{\partial f(x,y)}{\partial x}$ and $f_y(x,y)=\frac{\partial f(x,y)}{\partial y}$ (the two partial derivatives of the function, one with respect to x, the other with respect to y), you should then be able to see that the domains of those resulting functions are no more restrictive than the domain of the original function.

For example, when I look at your original function, I see that it has vertical asymptotes (places where one of the denominators goes to zero) at x=0 and at y=0. That means that the domain is

$x=[-\infty,0),(0,\infty]$
$y=[-\infty,0),(0,\infty]$

or, more concisely

$x \neq 0$
$y \neq 0$

If you compute the two partial derivatives, you should find their domains are the same or even less restrictive (i.e. x=0 and/or y=0 might be allowed as part of the domain for $f_x(x,y)$ or $f_y(x,y)$ ).

By the way, in case you're not familiar with partial derivatives, it's a pretty simple concept. Just treat all variables as constants except for the one with respect to which you're differentiating.

For example, if

$f(x,y)=2x^3y^4z$

then

$\frac{\partial f(x,y)}{\partial x} = 6x^2y^4z$

$\frac{\partial f(x,y)}{\partial y} = 8x^3y^3z$

and

$\frac{\partial f(x,y)}{\partial z} = 2x^3y^4$

Does that make sense?