Page 150 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 6] PARTIAL DERIVATIVES 141

@ @

ð cos Þ ð cos Þ
@ @ cos sin
@ðx; yÞ
@ @
ðcÞ ¼ ¼
sin cos
@ð ; Þ
ð sin Þ ð sin Þ
@ @
2
2
¼ ðcos þ sin Þ¼
(d)From Problem 6.43(b)we have, letting u ¼ , v ¼ ,
1
¼ 1 so that, using ðcÞ; ¼
@ðx; yÞ @ð ; Þ @ð ; Þ

@ð ; Þ @ðx; yÞ @ðx; yÞ
This can also be obtained by direct differentiation.
Note that from the Jacobians of these transformations it is clear why ¼ 0 (i.e., x ¼ 0, y ¼ 0) is a
singular point.
MEAN VALUE THEOREMS
6.40. Prove the mean value theorem for functions of two variables.
Let f ðtÞ¼ f ðx 0 þ ht; y 0 þ ktÞ. By the mean value theorem for functions of one variable,
0 0 < < 1
Fð1Þ¼ Fð0Þ¼ F ð Þ ð1Þ
If x ¼ x 0 þ ht, y ¼ y 0 þ kt,then FðtÞ¼ f ðx; yÞ,sothat by Problem 6.17,
0 0
F ðtÞ¼ f x ðdx=dtÞþ f y ðdy=dtÞ¼ hf x þ kf y and F ð Þ¼ hf x ðx 0 þ h; y 0 þ kÞþ kf y ðx 0 þ h; y 0 þ kÞ
where 0 < < 1. Thus, (1)becomes
f ðx 0 þ h; y 0 þ kÞ f ðx 0 ; y 0 Þ¼ hf x ðx 0 þ h; y 0 þ kÞþ kf y ðx 0 þ h; y 0 þ kÞ ð2Þ
where 0 < < 1as required.
Note that (2), which is analogous to (1)ofProblem 6.14 where h ¼ x, has the advantage of being
more symmetric (and also more useful), since only a single number is involved.

MISCELLANEOUS PROBLEMS
8 !
2 2
x y
>
xy ðx; yÞ 6¼ð0; 0Þ
<
2
x þ y 2 .
6.41. Let f ðx; yÞ¼
>
0 ðx; yÞ¼ ð0; 0Þ
:
Compute (a) f x ð0; 0Þ; ðbÞ f y ð0; 0Þ; ðcÞ f xx ð0; 0Þ; ðdÞ f yy ð0; 0Þ; ðeÞ f xy ð0; 0Þ; ð f Þ f yx ð0; 0Þ:
0
f x ð0; 0Þ¼ lim f ðh; 0Þ f ð0; 0Þ ¼ lim ¼ 0
ðaÞ
h!0 h h!0 h
0
f y ð0; 0Þ¼ lim f ð0; kÞ f ð0; 0Þ ¼ lim ¼ 0
ðbÞ
h!0 k k!0 k
If ðx; yÞ 6¼ð0; 0Þ,
( !) ! !
2
2
@ x y 2 4xy 2 x y 2
xy ¼ xy þ y
@x x þ y ðx þ y Þ x þ y
f x ðx; yÞ¼ 2 2 2 2 2 2 2
( !) ! !
2
2
@ x y 2 4xy 2 x y 2
xy ¼ xy þ x
f y ðx; yÞ¼ 2 2 2 2 2 2
@y x þ y 2 x þ y
ðx þ y Þ