Muradean said:
hum... that seems kind of vague, i just wanted to know why in some cases i cannot calculate the partial derivates by the the rules and i need to use the definition: let me ilustrate with an example... (Lets say i want to calculate the partial derivatives of f(x,y) and f(x,y) is:
| (y³) / (x⁴ + y²) if( x,y) is not 0
|
| 0 if (x,y) is 0
For any ##(x,y) \neq (0,0)## the function "behaves well", and so rules apply readily. Of course, the rules are derived from the definition, but are easier and faster to use than the definition itself.
However, at certain points the "rules" can become problematic; expect trouble whenever you try to divide by 0, or whenever you reach the boundary of the domain of a root-function for a root < 1; for example, trouble can occur at ##x = 0## if your function contains ##\sqrt{x}## or ##x^{3/4}##, or ##x^p## for positive ##p < 1##. In your example ##f(x,y)##, we can expect trouble when ##x = y = 0##, because we would then be dividing by 0
if we tried to use the formula in line 1. Of course, when ##(x,y) = (0,0)## we use line 2, giving ##f(0,0) = 1##, so in the actual ##f## we are not dividing by 0 because we switch to a different formula.
At a "troublesome" point, you need to fall back on the actual definition of the partial derivatives. By definition:
$$f_x(0,0) = \lim_{h \to 0} \frac{f(h,0) - f(0,0)}{h} = \lim_{h \to 0} \frac{0-0}{h} = 0,$$
and
$$f_y(0,0) = \lim_{h \to 0} \frac{f(0,h) - f(0,0)}{h} = \lim_{h \to 0}\frac{(h^3/h^2) - 0}{h} = \lim_{h \to 0} 1 = 1$$
