Page 139 - Schaum's Outline of Theory and Problems of Advanced Calculus

130 PARTIAL DERIVATIVES [CHAP. 6

!
3
2
2
2
2
@z
@ z @ @ x 3 ðx þ y Þð3x Þ ðx Þð2xÞ 2 3 1 2
2
@x @y ¼ @x @y ¼ @x x þ y 2 ¼ 2 2 2 ¼ 2 2 ¼ 1at ð1; 1Þ
ðx þ y Þ
The result can be written z xy ð1; 1Þ¼ 1:
Note:Inthis calculation we are using the fact that z xy is continuous at ð1; 1Þ (see remark at the end of
Problem 6.9).
6.13. If f ðx; yÞ is defined in a region r and if f xy and f yx exist and are continuous at a point of r, prove
that f xy ¼ f yx at this point.
Let ðx 0 ; y 0 Þ be the point of r. Consider
G ¼ f ðx 0 þ h; y 0 þ kÞ f ðx 0 ; y 0 þ kÞ f ðx 0 þ h; y 0 Þþ f ðx 0 ; y 0 Þ
Define (1) ðx; yÞ¼ f ðx þ h; yÞ f ðx; yÞ (2) ðx; yÞ¼ f ðx; y þ kÞ f ðx; yÞ
Then (3) G ¼ ðx 0 ; y 0 þ kÞ ðx 0 ; y 0 Þ (4) G ¼ ðx 0 þ h; y 0 Þ ðx 0 ; y 0 Þ
Applying the mean value theorem for functions of one variable (see Page 72) to (3) and (4), we have
(5) G ¼ k y ðx 0 ; y 0 þ 1 kÞ¼ kf f y ðx 0 þ h; y 0 þ 1 kÞ f y ðx 0 ; y 0 þ 1 kÞg 0 < 1 < 1
(6) G ¼ h x ðx 0 þ 2 h; y 0 Þ¼ hf f x ðx 0 þ 2 h; y 0 þ kÞ f x ðx 0 þ 2 h; y 0 Þg 0 < 2 < 1
Applying the mean value theorem again to (5) and (6), we have
(7) G ¼ hk f yx ðx 0 þ 3 h; y 0 þ 1 kÞ 0 < 1 < 1; 0 < 3 < 1
(8) G ¼ hk f xy ðx 0 þ 2 h; y 0 þ 4 kÞ 0 < 2 < 1; 0 < 4 < 1
From (7) and (8)we have

ð9Þ f yx ðx 0 þ 3 h; y 0 þ 1 kÞ¼ f xy ðx 0 þ 2 h; y 0 þ 4 kÞ
Letting h ! 0 and k ! 0in(9)we have, since f xy and f yx are assumed continuous at ðx 0 ; y 0 Þ,

f yx ðx 0 ; y 0 Þ¼ f xy ðx 0 ; y 0 Þ
as required. For example where this fails to hold, see Problem 6.41.

DIFFERENTIALS

6.14. Let f ðx; yÞ have continuous first partial derivatives in a region r of the xy plane. Prove that
f ¼ f ðx þ x; y þ yÞ f ðx; yÞ¼ f x x þ f y y þ 1 x þ 2 y
where 1 and 2 approach zero as x and y approach zero.
Applying the mean value theorem for functions of one variable (see Page 72), we have

ð1Þ f ¼f f ðx þ x; y þ yÞ f ðx; y þ yÞg þ f f ðx; y þ yÞ f ðx; yÞg
0 < 1 < 1; 0 < 2 < 1
¼ xf x ðx þ 1 x; y þ yÞþ yf y ðx; y þ 2 yÞ
Since, by hypothesis, f x and f y are continuous, it follows that
f x ðx þ 1 x; y þ yÞ¼ f x ðx; yÞþ 1 ; f y ðx; y þ 2 yÞ¼ f y ðx; yÞþ 2
where 1 ! 0, 2 ! 0as x ! 0 and y ! 0.
Thus, f ¼ f x x þ f y y þ 1 x þ 2 y as required.
Defining x ¼ dx; y ¼ dy,we have f ¼ f x dx þ f y dy þ 1 dx þ 2 dy:
We call df ¼ f x dx þ f y dy the differential of f (or z)orthe principal part of f (or z).

2
6.15. If z ¼ f ðx; yÞ¼ x y 3y, find (a) z; ðbÞ dz: ðcÞ Determine z and dz if x ¼ 4, y ¼ 3,
x ¼ 0:01, y ¼ 0:02. (d)How might you determine f ð5:12; 6:85Þ without direct computa-
tion?