Say you have a function of two variables, f(x,y). This function is differentiable at (X[sub]0[/sub], y[sub]0[/sub]) if it can be written as

Δf = f[sub]x[/sub](x[sub]0[/sub], y[sub]0[/sub])Δx + f[sub]y[/sub](x[sub]0[/sub], y[sub]0[/sub])Δx + ε[sub]1[/sub]Δx + ε[sub]2[/sub]Δy

where ε[sub]1[/sub] and ε[sub]2[/sub] are functions that go to zero as Δx and Δy go to zero.

My question is, how do you find those functions? I understand how ε is defined for a one variable function, but I’m not sure how it is defined for a two variable function. It seems broad to me. For instance, why not just make ε1 and ε2 the constant function 0? But then every function would be differentiable. there must be limits on the functions you can pick.

How can you use this definition to show that a function is differentiable? How are the “epsilon” functions chosen? could someone give an example using a simple function like x^2 + y^2 or sin(x)cos(y)

Thanks in advance, this has been bugging me. I know there are simpler ways to determine differentiability, but I’d like to know how you can use it directly.

Note that Δf means the change in f as (x, y) goes from (x[sub]0[/sub], y[sub]0[/sub]) to (x[sub]0[/sub]+Δx, y[sub]0[/sub]+Δy) - i.e.

Δf = f(x[sub]0[/sub]+Δx, y[sub]0[/sub]+Δy) – f(x[sub]0[/sub], y[sub]0[/sub])

You can work out what this expression looks like for any particular f.

You can also work out what f[sub]x[/sub](x[sub]0[/sub], y[sub]0[/sub]) and f[sub]y[/sub](x[sub]0[/sub], y[sub]0[/sub]) look like, and hence what f[sub]x[/sub](x[sub]0[/sub], y[sub]0[/sub])Δx + f[sub]y[/sub](x[sub]0[/sub], y[sub]0[/sub])Δy look like.

By matching up the two sides and seeing what’s “left over,” you get what the ε’s are.

Roughly, f is differentiable if the total differential of f is close to (or approaches) the change in f when the changes in x and y are close to (or approach) 0.

Example: f(x, y) = 3x² + xy – 5y.

Δf = f(x+Δx, y+Δy) – f(x, y) (for convenience I’m leaving off the subscripty 0’s)
= 3(x+Δx)² + (x+Δx)(y+Δy) - 5(y+Δy) – [3x² + xy – 5y]
which, after a little algebra, simplifies to
6xΔx + (Δx)² + xΔy + yΔx + ΔxΔy – 5Δy.

Motivated by the fact that f[sub]x[/sub] = 6x + y and f[sub]y[/sub] = x – 5, I rearrange this as
(6x + y)Δx + (x – 5)Δy + (Δx)² + ΔxΔy

This is indeed f[sub]x[/sub]Δx + f[sub]y[/sub]Δy + (something)Δx + (something)Δy,
where both of the somethings approach 0 as (Δx, Δy) approaches (0, 0).

Thanks, Thudlow! I didn’t think of writing out Δf. Not sure why that escaped me.

Whoops—just noticed this should have been 6xΔx + 3(Δx)² + xΔy + yΔx + ΔxΔy – 5Δy

Not that it affects the basic idea.