Page 137 - Schaum's Outline of Theory and Problems of Advanced Calculus

128 PARTIAL DERIVATIVES [CHAP. 6

Since the limit of the function depends on the manner of approach to ð0; 0Þ (i.e., the slope m of the line),
the function cannot be continuous at ð0; 0Þ.
Another method:
( 2 2 ) 2 ( 2 2 )
x y x x y
Since lim lim ¼ lim ¼ 1 and lim lim ¼ 1are not equal, lim f ðx; yÞ
2
2
x!0 y!0 x þ y 2 x!0 x 2 y¼0 x!0 x þ y 2 x!0
y!0
cannot exist. Hence, f ðx; yÞ cannot be continuous at ð0; 0Þ.
PARTIAL DERIVATIVES
2
2
6.8. If f ðx; yÞ¼ 2x xy þ y , find (a) @f =@x, and (b) @f =@y at ðx 0 ; y 0 Þ directly from the definition.

@f f ðx 0 þ h; y 0 Þ f ðx 0 ; y 0 Þ
¼ f x ðx 0 ; y 0 Þ¼ lim

ðaÞ
@x h!0 h
ðx 0 :y 0 Þ
2 2 2 2
¼ lim ½2ðx 0 þ hÞ ðx 0 þ hÞy 0 þ y 0 Š¼½2x 0 x 0 y 0 þ y 0 Š
h!0 h
2
4hx 0 þ 2h hy 0
¼ lim ¼ lim ð4x 0 þ 2h y 0 Þ¼ 4x 0 y 0
h!0 h h!0

@f f ðx 0 ; y 0 þ kÞ f ðx 0 ; y 0 Þ
¼ f y ðx 0 ; y 0 Þ¼ lim
ðbÞ
@y k!0 k
ðx 0 ;y 0 Þ
2 2 2 2
¼ lim ½2x 0 x 0 ðy 0 þ kÞþð y 0 þ kÞ Š ½2x 0 x 0 y 0 þ y 0 Š
k!0 k
kx 0 þ 2ky 0 þ k 2
¼ lim ¼ lim ð x 0 þ 2y 0 þ kÞ¼ x 0 þ 2y 0
k!0 k k!0
Since the limits exist for all points ðx 0 ; y 0 Þ,wecan write f x ðx; yÞ¼ f x ¼ 4x y, f y ðx; yÞ¼ f y ¼
x þ 2y which are themselves functions of x and y.
Note that formally f x ðx 0 ; y 0 Þ is obtained from f ðx; yÞ by differentiating with respect to x,keeping y
constant and then putting x ¼ x 0 ; y ¼ y 0 . Similarly, f y ðx 0 ; y 0 Þ is obtained by differentiating f with
respect to y,keeping x constant. This procedure, while often lucrative in practice, need not always
yield correct results (see Problem 6.9). It will work if the partial derivatives are continuous.

2 2
: Prove that (a) f x ð0; 0Þ and f y ð0; 0Þ both exist but
xy=ðx þ y Þðx; yÞ 6¼ð0; 0Þ
6.9. Let f ðx; yÞ¼
0 otherwise
that (b) f ðx; yÞ is discontinuous at 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; 0Þ f ð0; 0Þ ¼ lim ¼ 0
k!0 k k!0 k
mx 2 m
(b)Let ðx; yÞ! ð0; 0Þ along the line y ¼ mx in the xy plane. Then lim f ðx; yÞ¼ lim ¼
2 2
2
x!0 x!0 x þ m x 1 þ m 2
y!0
so that the limit depends on m and hence on the approach and therefore does not exist. Hence, f ðx; yÞ
is not continuous at ð0; 0Þ:
Note that unlike the situation for functions of one variable, the existence of the first partial
derivatives at a point does not imply continuity at the point.
2
3
2
y x y x xy 2
and f x ð0; 0Þ, f y ð0; 0Þ cannot be
Note also that if ðx; yÞ 6¼ð0; 0Þ, f x ¼ 2 2 , f y ¼ 2 2
2 2
ðx þ y Þ ðx þ y Þ
computed from them by merely letting x ¼ 0 and y ¼ 0. See remark at the end of Problem 4.5(b)
Chapter 4.
3 xy 2
6.10. If ðx; yÞ¼ x y þ e , find (a) x ; ðbÞ y ; ðcÞ xx ; ðdÞ yy ; ðeÞ xy ; ð f Þ yx .