1550251515-Classical_Complex_Analysis__Gonzalez_

Differentiation 315

but this is not an infinitesimal with respect to l~zl, since

e ~x~y

l~zl = ~x2 + ~y2

has no limit as ~x ---t 0 and ~y ---t 0, as can be seen by letting ~y = k ~x,

with k i= 0 an arbitrary constant. ·

The example shows, at the same time, that the existence of the partial

derivatives ux, Uy is not a sufficient condition for a function to be differen-

tiable at a point. This is not to say that the existence of Ux and Uy may

not suffice in some special cases. What is meant is that the mere existence

of ux, Uy at a point does not ensure in general the differentiability of u

at the same point.

For example, if u is of the form u(x,y) = g(x) + h(y), (x,y) ED, D

open, and we assume the existence of Ux = g'(x) and Uy = h'(y) at some

point (x, y) E D, then we have

~u = [g(x + ~x) - g(x)] + [h(y + ~y) - h(y)]

and by the definition of the ordinary derivative,

g(x+~x)-g(x) '()

~x = g x +s1

h(y + ~y) - h(y) = h'(y) + e2

~y

with e 1 ---t 0 as ~x ---t O, s 2 ---t 0 as ~y ---t 0. Hence we obtain

~u = [g'(x) + e1] ~x + [h'(y) + e2] ~y

= Ux ~x + Uy ~y + e1 ~x + e2 ~y

which shows that in this case u is differentiable at (x, y).
Sufficient conditions for the differentiability of a function u: D ---t R, D
open in R^2 , at a point (x, y) E D are given by the following theorems.

Theorem 6.3 Suppose that (1) Ux and Uy exist in a neighborhood
N 0 (x, y) C D, and (2) Ux and Uy are continuous at (x, y). Then u is
differentiable at (x, y).

Proof See any advanced calculus book.

Theorem 6.4 Suppose that (1) Ux exists in a neighborhood N 0 (x, y) C
D, (2) Ux is continuous at (x,y), and (3) Uy exists at (x,y). Then u is
differentiable at ( x, y).
Proof We have

~u = u(x + ~x,y + ~y)-u(x,y)