naggy said:
f(x,y) = (2xy)/(x2 + y2), 0 if (x,y) = (0,0)
Now I'm supposed to evaluate this at (0,0). I take the first partial derivative and I get 0/0 but when I use the definition of derivatives I get a whole number. Why the hell is this?
Surely you do not mean "I'm supposed to evaluate this at (0,0)". That's easy: it's 0!
As for the derivatives, I'm afraid I get exactly the opposite result to you.
The derivative with respect to x is taken for y fixed so it is the limit of [f(0+h,0)- f(0,0)]/h= 0. The derivative with respect to y is the same. (If you just "plug in" h=0 you get 0/0 but you know from Calculus I that that happens for
any derivative.) The derivative is 0 because f(h,0)= 0 for any h. 0/h= 0 which has limit 0 as h goes to 0. Of course, the same is true the partial derivative with respect to y.
In order that the "derivative" (really the "gradient"), as opposed to the partial derivatives, the limit of [f(x+h_1,y+h_2)- f(x,y)]/\sqrt{h_1^2+ h_2^2} must exist as h_1 and h_2 go to 0
independently. Since we are talking about (x,y)= (0,0) this is just the same as taking the limit of
\frac{2xy}{(x^2+ y^2)^{3/2}}
as (x, y) goes to (0, 0). (I've replaced h_1 and h_2 by x and y for convenience.)
Best way to find a limit like that is to convert to polar coordinates:
\frac{2(r cos(\theta))(r sin(\theta))}{r^3}= \frac{2 sin(\theta)cos(\theta)}{r}
which clearly does not exist as r goes to 0.
I suspect that your textbook is trying to convince you that "having partial derivatives" and "being differentiable" are not the same for functions of more than one variable. A function of several variables is "differentiable" at a point if and only if its partial derivatives are continuous at that point.
