Thomas Calculus 13th [Solutions]

996 Chapter 14 Partial Derivatives
69. Differentiating each equation implicitly gives 1 v ln v
x u u
u x and 0 ux
ln v u v
v x or
1
(ln u) v v
x u 1
u x 0 ln v ln v
vx v
u
ln (ln u)(ln v) 1
v (ln ) 0
u
u
v x v ux
ln v
u
v
v
u
70. Differentiating each equation implicitly gives 1 (2 x) xu
(2 y)
y u and 0 (2 x) x u y u or
1 2 y
2x
1
(2 x) xu
(2 y) yu
1
0 1 1 1
2x 0
xu and y 2x 2x
1
(2 x) x
0
2x
2 y 2x 4xy 2x 4xy
u
u y
2x 4xy 2x 4xy 2x 4xy 1 2 y
;
u
2x
1
2 2 y 2 1 2
next
1 1 1 y y
s x y s 2x x 2y 2x 2y
u u u 2x 4xy 1 2y 1 2y 1 2 y 1 2y
71.
0 if y 0
( , 0 ) ( , 0)
fx
( x, y) fx
( x, y) 0 for all points ( x, y ); at y
0 if y 0
0, f y ( x , 0) lim f x h f x
h 0
h
f ( x, h) 0 f ( x, h)
f ( x, h)
3
lim lim 0 because lim lim h
f ( x, h)
2
0 and lim lim h 0
h 0
h
h 0
h
h
h
h
h
h 0 h 0
h 0 h 0
2
3y
if y 0
f y ( x, y) ; f yx ( x, y) fxy
( x, y ) 0 for all points ( x, y)
2y
if y 0
f (0 h, y) f (0, y) f ( h, y) 0 f ( h, y)
72. At x 0, fx (0, y ) lim lim lim which does not exist because
h 0
h
h 0
h
h 0
h
1
f ( h, y)
2
lim lim h
f ( h, y) if x 0
h
0 and lim lim lim 1 f ( , ) 2 x ;
h
h
x x y
h h
h 0 h 0
h 0 h 0 h 0 h
2x
if x 0
0 if x 0
f y ( x, y) f y ( x, y) 0 for all points ( x, y ); f yx ( x, y ) 0 for all points ( x, y ), while
0 if x 0
fxy ( x , y ) 0 for all points ( x, y ) such that x 0.
f f
73. 2 x,
2 y,
x y
f
z
f
4z
2,
2
x
2
2
f
2,
2
y
2 2 2 2
f f f f
4 2 2 ( 4) 0
2 2 2 2
z x y z
f
f
74. 6 xz,
6 yz,
x
z
f
z
2 2 2
6z 3 x y ,
2
f
6 z,
2
x
2
f
6 z,
2
y
2
f
12z
2
z
2 2 2 f
f f
6z 6z 12z
0
2 2 2
x y z
75.
f
x
2y
2e
sin 2 x,
f
y
2 y
2e
cos 2 x,
2
x
f
2
2 y
4e
cos 2 x,
2
y
f
2
2 y
4e
cos 2x
2 2
f f 2 2
4e y cos 2x 4e y cos 2x
0
2 2
x y
f
76. x
x , 2 2
x y
f y
y x y
2 2 ,
2 2 2
f y x
2 2
2 2
x x y
,
2 2 2 2 2 2 2 2 2
f x y f f y x x y
2 2 2 2 2 2
2 2 2 2 2 2
y x y x y x y x y
0
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