Rigorously determining differentiability in multiple variables

[Gauss M.D.]

Homework Statement

Determine if

f(x,y) = ((x-y)⁴ +x³ +xy²)/(x²+y²)
[f(x,y = 0 @ (0,0)]

is differentiable at the origin.

Homework Equations

x = (0,0)

The Attempt at a Solution

A function is differentiable at x if f(x+Δx) - f(x) = AΔx + |Δx|R(x)

Where A are constant coefficients of the vector Δx, and R(Δx) → 0 as Δx → 0.

A couple of questions.

1) in order to solve this, should I set A = [∂f/∂x, ∂f/∂y] evaluated at (0,0)? My literature claims A just needs to be "some constant", not literally the partials evaluated at that point. A bit confused there.

2) Just to double check, if the first step is determining ∂f/∂x at a given point, I am free to set y=0 BEFORE calculating the partial wrt x, right?

 

[HallsofIvy]

Homework Statement

Determine if

f(x,y) = ((x-y)⁴ +x³ +xy²)/(x²+y²)
[f(x,y = 0 @ (0,0)]

is differentiable at the origin.

Homework Equations

x = (0,0)

The Attempt at a Solution

A function is differentiable at x if f(x+Δx) - f(x) = AΔx + |Δx|R(x)

Where A are constant coefficients of the vector Δx, and R(Δx) → 0 as Δx → 0.

A couple of questions.

1) in order to solve this, should I set A = [∂f/∂x, ∂f/∂y] evaluated at (0,0)? My literature claims A just needs to be "some constant", not literally the partials evaluated at that point. A bit confused there.

From the point of view of defining "differentiable", "A" just has to be some fixed vector. But then one can show that it is, in fact, the partials evaluated at that point.

2) Just to double check, if the first step is determining ∂f/∂x at a given point, I am free to set y=0 BEFORE calculating the partial wrt x, right?

Yes, in calculating the partial derivative, at a given point, you hold all the other variables constant.

In fact, an example of a function which has partial derivatives at a given point but is NOT differentiable there is f(x,y)= 0 if xy= 0, 1 otherwise. To find the partial derivative with respect to x, at (0, 0), we set y= 0 so xy= 0 for all x: f(x,0)= 0 for all x so has derivative 0. If y were allowed to have non-zero value, xy= 0 only for x= 0 so f(x,0)= 0 if x=0, 1 otherwise and is not differentiable.