given a function f: R^2 -> R, and a point (x,y) in R^2, state...?

Question

...precisely what it means for f to be:

(a) continuous at (x,y),
(b) differentiable at (x,y).

and if f is differentiable at (x,y), what is the derivative of f?

Answer 1

Jim was right about the definition of continuity, but was wrong about the definition of differentiability. We say the function f is differentiable at (x,y), if there is a linear functional L:R^2->R such that
|| f(u,v)-f(x,y)-L(u-x,v-y)||
divided by ||(u-x,v-y)||
goes to 0 as (u,v) goes to (x,y). Here, ||(a,b)|| is the length of the vector (a,b)=sqrt(a^2+b^2) and this is a two variable limit. We then say that the functional L is the total derivative of f at (x,y).

Now, if such an L exists, it is easy to show that the partial derivatives mentioned by Jim do exist and that
L(a,b)=f_x(x,y) a +f_y(x,y) b.
But even if both partial derivatives exist, it does not mean that the function is differentiable at the point. It does turn out that if the partial derivatives are themselves continuous, then f is differentiable, but that needs to be proved.

Answer 2

For (a), f(x,y) is continuous at (a,b) if for every \epsilon >0 there is a \delta >0 so that
|f(x,y)-f(a,b)|<\epsilon whenever \sqrt{(x-a)^2+(y-b)^2} < \delta.

For (b), The discussion begins with partial derivatives. Loosly speaking, the partial derivative of f, f_x, with respect to x is found by differentiating f with respect to f (IE., pretending y is constant). Similarly, the partial of f with respect to y, f_y, is found by differentiating f with respect to y (IE., pretending x is constant).

Then, the derivative of f at (a,b) is defined to be

\Delta f=f_x(a,b) \Delta x+f_y(a,b) \Delta y+\epsilon_1 \Delta x+\epsilon_2 \Delta y

where \epsilon_1 and \epsilon_2 are functions of \Delta x and \Delta y that have limit zero as (\Delta x, \Delta y) approaches (0,0).

You can find all the various rules in a standard calculus text.

Finally, to completely answer your question you need to give one more piece of information: what do you want to differentiate f with respect to? For example, if x and y are each functions of t (or s and t), you'll need to use the Chain rule and the problem will become more complex.

Answer 3

Domain: R^2
Range: R
(a)
So, y = x^2 which is a continuous function for all real values of x.

(b) dy/dx = 2x
So f(x) is always differentiable for all real values of x.

So the derivative of f is 2x.