Partial Derivative of f(x,y) at (0,0)

[Kuma]

Homework Statement

The question asks:

f(x,y) = 3xy+5y^3/[x^2+y^2] when (x,y) =! (0,0)
f(x,y) = 0 when (x,y) = (0,0)

what is df/dy at (0,0)?

Homework Equations

The Attempt at a Solution

I'm not sure what the answer is. At 0,0 f(x,y) is 0, so it's simply a point and the function is not continuous at the point, therefore df/dy doesn't exist?

 

[Ray Vickson]

Homework Statement

The question asks:

f(x,y) = 3xy+5y^3/[x^2+y^2] when (x,y) =! (0,0)
f(x,y) = 0 when (x,y) = (0,0)

what is df/dy at (0,0)?

Homework Equations

The Attempt at a Solution

I'm not sure what the answer is. At 0,0 f(x,y) is 0, so it's simply a point and the function is not continuous at the point, therefore df/dy doesn't exist?

There are functions for which all partial derivatives exist at a point of discontinuity, so lack of continuity cannot be relied upon in this question.

RGV

 

[Kuma]

So then it must be 0? It's the partial derivative of a point.

 

[SammyS]

Homework Statement

The question asks:

f(x,y) = 3xy+5y^3/[x^2+y^2] when (x,y) =! (0,0)
f(x,y) = 0 when (x,y) = (0,0)

what is df/dy at (0,0)?

Homework Equations

The Attempt at a Solution

I'm not sure what the answer is. At 0,0 f(x,y) is 0, so it's simply a point and the function is not continuous at the point, therefore df/dy doesn't exist?

Can you show that f is not continuous at (0,0) ?

 

[vela]

Don't get confused by how the function is defined. It doesn't make sense to talk about the "partial derivative of a point." You find the partial derivative of a function at a point.

Go back to the basic definition. You want to find
\left.\frac{\partial f}{\partial y}\right|_{(x_0,y_0)} = \lim_{h \to 0} \frac{f(x_0,y_0+h)-f(x_0,y_0)}{h}when (x₀,y₀)=0. You need to find the limit or show it doesn't exist.