This doesnt seem right, is it? Partial derivitive at x,y = (0,0)?

Question

Consider f(x,y) = ((x^2)y)/((x^2)+(y^2)) (x,y) not equal (0,0)
0 (x,y) = (0,0)

find partial deriviatives at (0,0) using limits.

Ok.....heres my work....

fx(x,y) = [f(x+h,y) - (f(x,y))]/h --> after subing i get

fx(x,y) = ([0/h^2] - [0/0]) / h in turns gives me zero?


now y

fy(x,y) = [f(x,y+h) - f(x,y)] / h --> in turn gives me zero.

To me this doesnt seem right, am i right in assuming that 0/h^2 will give me zero. I know 0/0 is undefined but as we approach it are we approching 1 or zero or even infinity.

Answer 1

The only problem in your derivation is that you are evaluating f(0, 0) as 0/0. This is not the case, the second part of the definition is that f(x, y) = 0 when (x, y) = (0, 0). Nonetheless, your result is correct: f_x(0, 0) = [h→0]lim (0/h² - 0)/h = 0. Similarly with f_y(0, 0).

This makes sense, because the restriction of f(x, y) to the x-axis is a constant function (i.e. f(x, 0) = 0 for all x), so of course the partial derivative of it w.r.t. x is constant. Similarly, if you restrict f(x, y) to the y-axis, you also get a constant function.

Note that f(x, y) is not differentiable in the 2-variable sense at (0, 0) (there is no tangent plane at that point), so the multivariate chain rule will NOT hold, and the directional derivatives won't be related to the partial derivatives. This helps to highlight an important difference between the partial and total derivates - it is possible for all partial derivatives to be defined and for the function to still not be differentiable at a point.

Answer 2

You will not like my answer...

Let say I do like you. Take your argument and correct it like:

fx(x,y) =lim as h->0 [f(x+h,y) - (f(x,y))]/h --> after subing i get

fx(x,y) = lim as h->0 ([0/h^2] - [0]) / h

I replaced f(0,0) by 0, not by 0/0 wich is wrong. f(0,0)=0.

Written like that, the expression is 0 even before taking the limit. So, when taking the limit of 0, one gets 0.

But, the problem is that you take the limit from one direction only (from the direction (h,0)). But you could take the derivative from another direction (let say (h,h) and take h->0).

Unfortunately, it would be very difficult in this small space to explain all the details but you need to consider several direction and if the derivative is not the same by considering other directions, then it mean that the derivative does not exist. This is exactly what is happening in your case. Good luck!

Answer 3

ARG! Or rather Argand. I think this nasty function needs to have other guns trained on it before doing the limiting process.

It's simply MADE for the substitutions:
x = r cos (theta), y = r sin (theta).

Then f(x,y) ---> g(r,theta) = r^3 cos^2(theta) sin(theta)/r^2, i.e.

f(x,y) ---> g(r,theta) = r cos^2(theta) sin(theta).

Now for the evaluation/limiting processes.

Along the x-axis, this fn is always (1) finite and (2) zero (sin [theta] = 0); nothing funny happens as x (now r) tends to zero.

So f_x(0,0) = 0.

Along the y-axis, the same fn is again always finite and zero, with nothing funny going on; thus f_y(0,0) = 0 also.

So, nasty though this function certainly appeared at first sight, we managed to tame it and induce good behaviour. "Our mission would seem to be accomplished, Captain."

[ADDED THOUGHT: It's worth while trying to visualize the 3-D surface given by z = f(x,y) = g(r, theta). This surface simply passes through the original x- and y-axes, so nothing odd happens as one goes through the origin in any of the four (plus or minus) x- or y-directions along these axes. However, the situation is very different in ANY OTHER directions! The surface RISES linearly away from the origin in the 1st and 3rd quadrants, and FALLS in the 2nd and 4th quadrants. This means that EXCEPT ALONG THE ORIGINAL AXIS DIRECTIONS, there's a finite angle between the slopes as you pass through the origin. Thus, the function is ONLY DIFFERENTIABLE ALONG THE x- AND y-AXES. DAMNED CUNNING! I congratulate whoever thought up this simple but fiendishly behaved function.]

Live long and prosper.