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6.10. If  x; y x3 y exy , nd
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PARTIAL DERIVATIVES
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2 2 2 @ @ x3 y exy 3x2 y exy y2 3x2 y y2 exy @x @x 2 2 2 @ @ x3 y exy x3 exy 2xy x3 2xy exy @y @y
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  2 2 2 @2  @ @ @ 3x2 y y2 exy 6xy y2 exy y2 6xy y4 exy 2 @x @x @x @x
2 2 2 @ @2  @ @ 2xy x3 2xy exy 0 2xy exy exy @y @y @y2 @y
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2xy exy 2xy exy 2x 4x2 y2 exy 2x exy e xy
  2 2 2 @2  @ @ @ 3x2 y y2 exy 3x2 y2 exy 2xy exy 2y @y @x @y @x @y 3x2 2xy3 exy 2y exy
yx
  2 2 2 @2  @ @ @ x3 2xy exy 3x2 2xy exy y2 exy 2y @x @y @x @y @x 3x2 2xy3 exy 2y exy
Note that xy yx in this case. This is because the second partial derivatives exist and are continuous for all x; y in a region r. When this is not true we may have xy 6 yx (see Problem 6.41, for example).
6.11. Show that U x; y; z x2 y2 z2 1=2 satis es Laplace s partial di erential equation @2 U @2 U @2 U 2 2 0. @x2 @y @z
We assume here that x; y; z 6 0; 0; 0 . Then @U 1 x2 y2 z2 3=2 2x x x2 y2 z2 3=2 2 @x @2 U @ x x2 y2 z2 3=2 x 3 x2 y2 z2 5=2 2x x2 y2 z2 3=2 1 2 @x @x2 2 2 2 2 3x x y z 2x2 y2 z2 2 2 2 2 z2 5=2 2 z2 5=2 x y x y2 z2 5=2 x y @2 U 2y2 x2 z2 2 ; 2 @y x y2 z2 5=2 @2 U 2z2 x2 y2 2 : 2 @x x y2 z2 5=2
Similarly
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@2 U @2 U @2 U 2 2 0: @x2 @y @z
y @2 z at 1; 1 . 6.12. If z x2 tan 1 , nd x @x @y
@z 1 @  y x2 1 x3 x2 x2 2 2 2 @y x 2 x @y x y x y2 1 y=x
PARTIAL DERIVATIVES
[CHAP. 6
!   @2 z @ @z @ x3 x2 y2 3x2 x3 2x 2 3 1 2 1 at 1; 1 2 2 @x @y @x @y @x x y 22 x2 y2 2 The result can be written zxy 1; 1 1: Note: In this calculation we are using the fact that zxy is continuous at 1; 1 (see remark at the end of Problem 6.9).
6.13. If f x; y is de ned in a region r and if fxy and fyx exist and are continuous at a point of r, prove that fxy fyx at this point.
Let x0 ; y0 be the point of r. Consider G f x0 h; y0 k f x0 ; y0 k f x0 h; y0 f x0 ; y0 De ne Then (1) (3)  x; y f x h; y f x; y G  x0 ; y0 k  x0 ; y0 (2) (4) x; y f x; y k f x; y G x0 h; y0 x0 ; y0 0 < 1 < 1 0 < 2 < 1
Applying the mean value theorem for functions of one variable (see Page 72) to (3) and (4), we have (5) (6) G ky x0 ; y0 1 k kf fy x0 h; y0 1 k fy x0 ; y0 1 k g G h
x x0
2 h; y0 hf fx x0 2 h; y0 k fx x0 2 h; y0 g G hk fyx x0 3 h; y0 1 k G hk fxy x0 2 h; y0 4 k 0 < 1 < 1; 0 < 3 < 1 0 < 2 < 1; 0 < 4 < 1
Applying the mean value theorem again to (5) and (6), we have (7) (8)
From (7) and (8) we have 9 fyx x0 3 h; y0 1 k fxy x0 2 h; y0 4 k Letting h ! 0 and k ! 0 in (9) we have, since fxy and fyx are assumed continuous at x0 ; y0 , fyx x0 ; y0 fxy x0 ; y0 as required. For example where this fails to hold, see Problem 6.41.
DIFFERENTIALS 6.14. Let f x; y have continuous rst partial derivatives in a region r of the xy plane. f f x x; y y f x; y fx x fy y 1 x 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 1 f f f x x; y y f x; y y g f f x; y y f x; y g x fx x 1 x; y y y fy x; y 2 y 0 < 1 < 1; 0 < 2 < 1
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